Rd Sharma Xi 2020 _volume 1 Solutions for Class 12 Commerce Maths Chapter 21 Some Special Series are provided here with simple step-by-step explanations. These solutions for Some Special Series are extremely popular among Class 12 Commerce students for Maths Some Special Series Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2020 _volume 1 Book of Class 12 Commerce Maths Chapter 21 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2020 _volume 1 Solutions. All Rd Sharma Xi 2020 _volume 1 Solutions for class Class 12 Commerce Maths are prepared by experts and are 100% accurate.

Page No 21.10:

Question 1:

13 + 33 + 53 + 73 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2n-13

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=k=1nTk      = k=1n2k-13      = k=1n8k3-1-6k2k-1      = k=1n8k3-1-12k2+6k      = k=1n8k3-1-12k2+6k      =8k=1nk3-k=1n1-12k=1nk2+6k=1nk      =8n2n+124-n-12nn+12n+16+6 nn+1 2     =2n2n+12-n-2nn+12n+1+3nn+1     =nn+12nn+1-22n+1+3-n     =nn+12n2-2n+1-n     =n2n3-2n2+n+2n2-2n+1-1     =n2n3-n     =n22n2-1

Page No 21.10:

Question 2:

22 + 42 + 62 + 82 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2n2

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=k=1nTk      = k=1n2k2      = k=1n4k2      = 4k=1nk2      = 4nn+12n+16     =2n3n+12n+1

Page No 21.10:

Question 3:

1.2.5 + 2.3.6 + 3.4.7 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=nn+1n+4    =nn2+5n+4    =n3+5n2+4n

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=k=1nk3+5k=1nk2+4k=1nk Sn=k=1nk3+5k=1nk2+4k=1nk Sn=n2n+124+5nn+12n+16+ 4nn+1 2Sn=n2n+124+5nn+12n+16+2nn+1Sn=nn+12nn+12+52n+13+4Sn=nn+12n2+n2+10n+53+4Sn=nn+123n2+3n+20n+10+246Sn=nn+1123n2+23n+34

Page No 21.10:

Question 4:

1.2.4 + 2.3.7 +3.4.10 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=nn+13n+1=n3n2+4n+1=3n3+4n2+n

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=3k=1nk3+4k=1nk2+k=1nk Sn=3k=1nk3+4k=1nk2+k=1nk Sn=3n2n+124+4nn+12n+16+ nn+1 2Sn=3n2n+124+2nn+12n+13+nn+12Sn=nn+123nn+12+42n+13+1Sn=nn+123n2+3n2+8n+43+1Sn=nn+129n2+9n+16n+8+66Sn=nn+1129n2+25n+14

Page No 21.10:

Question 5:

1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:
Tn=1+2+3+4+5+...+n=nn+12=n2+n2

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=k=1nk2+k2Sn=12k=1nk2+kSn=12nn+12n+16+nn+12Sn=nn+142n+13+1Sn=nn+142n+43Sn=nn+12n+412Sn=nn+1n+26

Page No 21.10:

Question 6:

1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=nn+1=n2+n

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=k=1nk2+kSn=k=1nk2+k=1nkSn=nn+12n+16+nn+12Sn=nn+122n+13+1Sn=nn+122n+43Sn=nn+12n+46Sn=nn+1n+23

Page No 21.10:

Question 7:

3 × 12 + 5 ×22 + 7 × 32 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2n+1n2=2n3+n2

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=k=1n2k3+k2Sn=2k=1nk3+k=1nk2Sn=2n2n+124+nn+12n+16Sn=n2n+122+nn+12n+16Sn=nn+12nn+1+2n+13Sn=nn+123n2+3n+2n+13Sn=nn+123n2+5n+13Sn=nn+163n2+5n+1

Page No 21.10:

Question 8:

Find the sum of the series whose nth term is:
(i) 2n2 − 3n + 5
(ii) 2n3 + 3n2 − 1
(iii) n3 − 3n
(iv) n (n + 1) (n + 4)
(v) (2n − 1)2

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

(i)
Tn=2n2-3n+5

Let Sn be the sum of n terms of the given series.

Now,
Sn=k=1nTk 

Sn=k=1n2k2-3k+5Sn=2k=1nk2-3k=1nk + k=1n5Sn=2nn+12n+16-3nn+12+5nSn=2nn+12n+1-9nn+1+30n6Sn=2n2+2n2n+1-9n2-9n+30n6Sn=4n3+4n2+2n2+2n-9n2-9n+30n6Sn=4n3-3n2+23n6Sn=n4n2-3n+236

(ii)
Tn=2n3+3n2-1

Let Sn be the sum of n terms of the given series.

Now,
Sn=k=1nTk 

Sn=k=1n2k3+3k2-1Sn=2k=1nk3+3k=1nk2 - k=1n1Sn=2n2n+124+3nn+12n+16-nSn=n2n+122+nn+12n+12-nSn=n2n+12+nn+12n+1-2n2Sn=n2n2+1+2n+n2+n2n+1-2n2Sn=n4+n2+2n3+2n3+n2+2n2+n-2n2Sn=n4+4n2+4n3-n2Sn=nn3+4n+4n2-12

(iii)
Tn=n3-3n

Let Sn be the sum of n terms of the given series.

Now,
Sn=k=1nTk 

Sn=k=1nk3-3kSn=k=1nk3-k=1n3kSn=n2n+124-3+32+33+34+...+3nSn=n2n+124-33n-13-1Sn=n2n+124-323n-1

(iv)
Tn=nn+1n+4=n2+nn+4=n3+5n2+4n

Let Sn be the sum of n terms of the given series.

Now,
Sn=k=1nTk 

Sn=k=1nk3+5k2+4kSn=k=1nk3 + 5k=1nk2 + 4k=1nkSn=n2n+124+5nn+12n+16+4nn+12Sn=nn+12nn+12+52n+13+4Sn=nn+1123nn+1+102n+1+24Sn=nn+1123n2+23n+34


(v)
Tn=2n-12

Let Sn be the sum of n terms of the given series.

Now,
Sn=k=1nTk 

Sn=k=1n2k-12Sn=k=1n4k2+1-4k Sn=4k=1nk2+1k=1n-4k=1nk Sn=4nn+12n+16+n-4nn+12Sn=nn+1242n+13-4+nSn=nn+128n+4-123+nSn=nn+128n-83+nSn=4nn+1n-13+nSn=n4n+4n-1+3n3Sn=n34n2+4n-4n-4+3Sn=n34n2-1Sn=n32n-12n+1

Page No 21.10:

Question 9:

Find the 20th term and the sum of 20 terms of the series 2 × 4 + 4 × 6 + 6 × 8 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2n2n+2=4n2+4n

For n = 20, we have:

T20=4n2+4n      =4202+420      =1600+80      =1680

Therefore, the 20th term of the given series is 1680.

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=k=1n4k2+4kSn=4k=1nk2+4k=1nk

For n = 20, we have:

S20=4k=120k2+4k=120kS20=42021416+420212S20=40741+4021S20=11480+840 =12320

Hence, the sum of the first 20 terms of the series is 12320.



Page No 21.18:

Question 1:

3 + 5 + 9 + 15 + 23 + ...

Answer:

Let Tn be the nth term and Sn be the sum to n terms of the given series.

Thus, we have:

Sn=3+5+9+15+23+... +Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     3+5+9+15+23+... +Tn-1+Tn   ...(2)

On subtracting (2) from (1), we get:

      Sn    =  3+5+9+15+23+... +Tn-1+Tn      Sn    =        3+5+9+15+23+... +Tn-1+Tn               0      = 3+2+4+6+8+... +Tn-Tn-1-Tn

The sequence of difference of successive terms is 2, 4, 6, 8,...
We observe that it is an AP with common difference 2 and first term 2.

Thus, we have:

3+n-124+n-22-Tn=03+n-122n=Tn3+nn-1=Tn

Now,
 Sn=k=1nTk  Sn=k=1n3+kk-1  Sn=k=1nk2+k=1n3-k=1nkSn=nn+12n+16+3n-nn+12Sn=n3n+12n+12+9-32n+1Sn=nn2+83

Page No 21.18:

Question 2:

2 + 5 + 10 + 17 + 26 + ...

Answer:

Let Tn be the nth term and Sn be the sum of n terms of the given series.

Thus, we have:

Sn=2+5+10+17+26+... +Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     2+5+10+17+26+... +Tn-1+Tn      ...(2)

On subtracting (2) from (1), we get:

Sn= 2+5+10+17+26+... +Tn-1+TnSn=       2+5+10+17+26+... +Tn-1+Tn       0=   2+3+5+7+9+... +Tn-Tn-1-Tn

The sequence of difference of successive terms is 3, 5, 7, 9,...

We observe that it is an AP with common difference 2 and first term 3.

Thus, we have:

2+n-126+n-22-Tn=02+n2-1=Tnn2+1=Tn

Now,
 Sn=k=1nTk  Sn=k=1nk2+1  Sn=k=1nk2+k=1n1Sn=nn+12n+16+nSn=nn+12n+1+6n6Sn=n2n2+3n+76

Page No 21.18:

Question 3:

1 + 3 + 7 + 13 + 21 + ...

Answer:

Let Tn be the nth term and Sn be the sum of n terms of the given series.

Thus, we have:

Sn=1+3+7+13+21+... +Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     1+3+7+13+21+... +Tn-1+Tn       ...(2)

On subtracting (2) from (1), we get:

Sn=1+3+7+13+21+... +Tn-1+TnSn=      1+3+7+13+21+... +Tn-1+Tn0 = 1+2+4+6+8+.. +Tn-Tn-1-Tn=0

The sequence of difference of successive terms is 2, 4, 6, 8,...

We observe that it is an AP with common difference 2 and first term 2.

Thus, we have:

1+n-124+n-22-Tn=01+n2-n=Tnn2-n+1=Tn

Now,
 Sn=k=1nTk  Sn=k=1nk2-k+1  Sn=k=1nk2+k=1n1-k=1nkSn=nn+12n+16+n-nn+12Sn=nn+122n-23+nSn=nn2-1+33Sn=nn2+23

Page No 21.18:

Question 4:

3 + 7 + 14 + 24 + 37 + ...

Answer:

Let Tn be the nth term and Sn be the sum of n terms of the given series.

Thus, we have:

Sn=3+7+14+24+37+ ... +Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     3+7+14+24+37+ ... +Tn-1+Tn      ...(2)

On subtracting (2) from (1), we get:

Sn=3+7+14+24+37+ ... +Tn-1+TnSn=      3+7+14+24+37+ ... +Tn-1+Tn    0  = 3+4+7+10+13+... +Tn-Tn-1-Tn

The sequence of difference of successive terms is 4, 7, 10, 13,...

We observe that it is an AP with common difference 3 and first term 4.

Thus, we have:

3+n-128+n-23-Tn=03+n-123n+2-Tn=03n2-n+42=Tn32n2-n2+2=Tn

Now,
 Sn=k=1nTk  Sn=k=1n32k2-k2+2  Sn=32k=1nk2+k=1n2-12k=1nkSn=nn+12n+14+2n-nn+14Sn=nn+12n+8n4Sn=n+12n2+8n4Sn=n2nn+1+4Sn=n2n2+n+4

Page No 21.18:

Question 5:

1 + 3 + 6 + 10 + 15 + ...

Answer:

Let Tn be the nth term and Sn be the sum of n terms of the given series.

Thus, we have:

Sn=1+3+6+10+15+... +Tn-1+Tn       ....(1)

Equation (1) can be rewritten as:

Sn=     1+3+6+10+15+ ... +Tn-1+Tn            ...(2)

On subtracting (2) from (1), we get:

Sn=1+3+6+10+15+... +Tn-1+TnSn=      1+3+6+10+15+... +Tn-1+Tn 0 =1+2+3+4+5+... +Tn-Tn-1-Tn

The sequence of difference of successive terms is 2, 3, 4, 5,...

We observe that it is an AP with common difference 1 and first term 2.

Thus, we have:

1+n-124+n-21-Tn=01+n-12n+2-Tn=0n2+n2=Tn

Now,
 Sn=k=1nTk  Sn=k=1nk2+k2  Sn=12k=1nk2+12k=1nkSn=nn+12n+112+nn+14Sn=nn+142n+13+1Sn=nn+142n+43Sn=nn+12n+23Sn=nn+1n+26

Page No 21.18:

Question 6:

1 + 4 + 13 + 40 + 121 + ...

Answer:

Let Tn be the nth term and Sn be the sum to n terms of the given series.

Thus, we have:

Sn=1+4+13+40+121+...+Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     1+4+13+40+121+...+Tn-1+Tn      ...(2)

On subtracting (2) from (1), we get:

    Sn=1+4+13+40+121+...+Tn-1+Tn    Sn=      1+4+13+40+121+...+Tn-1+Tn            0 =1+3+9+27+81+... +Tn-Tn-1-Tn

The sequence of difference between successive terms is 3, 9, 27, 81,...

We observe that it is a GP with common ratio 3 and first term 3.

Thus, we have:

1+33n-1-13-1-Tn=01+3n-32-Tn=03n2-12-Tn=03n2-12=Tn

 Sn=k=1nTk  Sn=k=1n3k2-12Sn=12k=1n3k-12k=1n1Sn=123+32+33+34+35+...+3n -n2Sn=1233n-12-n2Sn=3n+1-34-n2Sn=3n+1-3-2n4

Page No 21.18:

Question 7:

4 + 6 + 9 + 13 + 18 + ...

Answer:

Let Tn be the nth term and Sn be the sum of n terms of the given series.

Thus, we have:

Sn=4+6+9+13+18+...+Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     4+6+9+13+18+...+Tn-1+Tn       ...(2)

On subtracting (2) from (1), we get:

    Sn= 4+6+9+13+18+...+Tn-1+Tn    Sn=       4+6+9+13+18+...+Tn-1+Tn    0   =4+2+3+4+5+6+... +Tn-Tn-1-Tn

The sequence of difference between successive terms is 2, 3, 4, 5,...

We observe that it is an AP with common difference 1 and first term 2.

Now,

4+n-124+n-21-Tn=04+n-12n+2-Tn=04+n2+n2-1-Tn=0n22+n2+3=Tn

 Sn=k=1nTk  Sn=k=1nk22+k2+3        =12k=1nk2+12k=1nk+k=1n3        =nn+12n+12×6+nn+12×2+3n        =n2n2+3n+1+3n+3+3612        =n122n2+6n+40        =n6n2+3n+20

Page No 21.18:

Question 8:

2 + 4 + 7 + 11 + 16 + ...

Answer:

Let Sn be the sum of n terms and Tn be the nth term of the given series.

Thus, we have:

Sn=     2+4+7+11+16+...+Tn-1+Tn            ...(1)

Equation (1) can be rewritten as:

Sn=2+4+7+11+16+...+Tn-1+Tn            ...(2)

On subtracting (2) from (1), we get:

    Sn=     2+4+7+11+16+...+Tn-1+Tn    Sn=           2+4+7+11+16+...+Tn-1+Tn    -               -  -  -  -     -            -          -                     0  =   2+2+3+4+5+6+... +Tn-Tn-1-Tn

2+n-124+n-21-Tn=02+n-12n+2-Tn=02+n2+n2-1-Tn=0n22+n2+1=Tn

 Sn=k=1nTk  Sn=k=1nk22+k2+1        =12k=1nk2+12k=1nk+k=1n1        =nn+12n+112+nn+14+n        =n2n2+3n+1+3n+3+1212        =n122n2+6n+16        =n6n2+3n+8

Page No 21.18:

Question 9:

11.4+14.7+17.10+...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=1(3n-2) (3n+1)

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=k=1n13k-23k+1   =13k=1n13k-2-13k+1   =13k=1n13k-2-13k=1n13k+1   =131+14+17+110+...+13n-2-14+17+110+...+13n-2+13n+1   =131-13n+1   =n3n+1      

Page No 21.18:

Question 10:

11.6+16.11+111.14+114.19+...+1(5n-4) (5n+1)

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=1(5n-4) (5n+1)

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1n15k-45k+1    =15k=1n15k-4-15k+1    =15k=1n15k-4-15k=1n15k+1    =151+16+111+116+...+15n-4-16+111+116+...+15n-4+15n+1    =151-15n+1    =n5n+1      

Page No 21.18:

Question 1:

The sum to n terms of the series 11+3+13+5+15+7+....+.... is
(a) 2n+1

(b) 122n+1

(c) 2n+1-1

(d) 122n+1-1

Answer:

(d) 122n+1-1

Let Tn be the nth term of the given series.

Thus, we have:
Tn=12n-1+2n+1=2n+1-2n-12

Now,

Let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=k=1nTk      = k=1n2k+1-2k-12     =12 k=1n2k+1-2k-1     = 123-1+5-3+7-5+...+2n+1-2n-1     = 12-1+2n+1     = 122n+1-1



Page No 21.19:

Question 2:

The sum of the series 1log2 4+1log4 4+1log8 4+....+1log2n4 is
(a) n (n+1)2

(b) n (n+1) (2n+1)12

(c) n (n+1)4

(d) none of these

Answer:

(c) n (n+1)4

Let Sn=1log2 4+1log4 4+1log8 4+...+1log2n4 

Sn=log2log4+log4log4+log8log4+...+log2nlog4Sn=log2log4+log22log4+log23log4+...+log2nlog4  Sn=log2log4+2 log2log4+3 log2log4+...+n log2log4  Sn=log2log41+2+3+...+n Sn=log412log41+2+3+...+n Sn=12log4log41+2+3+...+n Sn=121+2+3+...+n Sn=nn+14

Page No 21.19:

Question 3:

The value of r=1n(2r-1) a+1br is equal to

(a) an2+bn-1-1bn-1 (b-1)

(b) an2+bn-1bn (b-1)

(c) an3+bn-1-1bn (b-1)

(d) none of these

Answer:

(b) an2+bn-1bn (b-1)

We have:
r=1n(2r-1) a+1br=r=1n2ra-a+1br=r=1n2ar-r=1na+r=1n1br=ann+1-an+1-bn1-bbn=an2+bn-1b-1bn

Page No 21.19:

Question 4:

If ∑ n = 210, then ∑ n2 =
(a) 2870
(b) 2160
(c) 2970
(d) none of these

Answer:

(a) 2870

Given:
n = 210

nn+12=210n2+n-420=0n-20n+21=0n=20                          n>0

Now,
n2=nn+12n+16n(n+1)2×(2n+1)3210 × 41370 × 412870

Page No 21.19:

Question 5:

If Sn = r=1n1+2+22+... Sum to r terms2r, then Sn is equal to
(a) 2nn − 1

(b) 1-12n

(c) n-1+12n

(d) 2n − 1

Answer:

(c) n-1+12n

We have:
Sn = r=1n1+2+22+...sum to r terms2r
Sn=r=1n12r-12rSn=r=1n1-12rSn=n-r=1n12rSn=n- 121 - 12n 1-12Sn=n- 1-12nSn=n-1+12n

Page No 21.19:

Question 6:

If 1+1+22+1+2+33+.... to n terms is S, then S is equal to

(a) n (n+3)4

(b) n (n+2)4

(c) n (n+1) (n+2)6

(d) n2

Answer:

a n (n+3)4

Let Tn be the nth term of the given series.

Thus, we have:

Tn=1+2+3+4+5+...+nn=nn+12n=n2+12

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn= k=1nk2+12 Sn= k=1nk2+n2 Sn= nn+14+n2 Sn= n2n+12+1 Sn= n2n+32 Sn= nn+34

Page No 21.19:

Question 7:

Sum of n terms of the series 2+8+18+32+ .... is
(a) n (n+1)2

(b) 2n (n + 1)

(c) n (n+1)2

(d) 1

Answer:

(c) n (n+1)2

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2×n2=n2

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=2 k=1nk Sn= 2nn+12 Sn= nn+12

Page No 21.19:

Question 8:

The sum of 10 terms of the series 2+6+18+.... is
(a) 121 (6+2)

(b) 243 (3+1)

(c) 1213-1

(d) 242 (3-1)

Answer:

(a) 121 (6+2)

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2×3n-1=23n-1

Now, let S10 be the sum of 10 terms of the given series.

Thus, we have:

 S10=2 k=1103k-1 S10= 21+3+32+...+39 S10= 2310-13-1 S10n= 235-13-13+13+1 S10= 2235-13+1 S10= 122426+2 S10= 121 6+2

Page No 21.19:

Question 9:

The sum of the series 12 + 32 + 52 + ... to n terms is
(a) n (n+1) (2n+1)2

(b) n (2n-1) (2n +1)3

(c) (n-1)2 (2n+1)6

(d) (2n+1)33

Answer:

(b) n (2n-1) (2n +1)3

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2n-12=4n2+1-4n

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn= k=1n4k2+1-4k Sn= 4   k=1   nk2 + 1k=1n - 4k=1nk Sn= 4nn+12n+1 6+n- 4nn+12 Sn= 2nn+12n+1 3+n- 2nn+1 Sn= n2n+12n+1 3+1- 2n+1 Sn= n32n+22n+1 +3 -6n+1 Sn= n34n2-1 Sn= n2n-12n+1 3

Page No 21.19:

Question 10:

The sum of the series 23+89+2627+8081+..... to n terms is
(a) n-12(3-n-1)

(b)n-12(1-3-n)

(c) n+12(3n-1)

(d) n-12(3n-1)

Answer:

(b) n-12(1-3-n)

Let Tn be the nth term of the given series.

Thus, we have:

Tn=3n-13n=1-13n

Now,

Let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=k=1nTk      = k=1n1-13k     = k=1n1- k=1n13k     = n-13+132+133+...+13n     = n-131-13n1-13     = n-121-13n     = n-121-3-n

Page No 21.19:

Question 11:

Let Sn denote the sum of the cubes of the first n natural numbers and Sn denote the sum of the first natural numbers. Then r=1n SrSr equals
(a) nn+1n+26

(b) nn+12

(c) n2+3n+22

(d) None of these

Answer:

Let Sn denote the sum of the cubes of the first n natural numbers and Sn denotes the sum of the first n natural number 1.
Then r=1n SrSr=? 
 Since Sn= nn+122   ; formula for sum of cubes. 

Sn=nn+12        ; formula form sum of first n natural numbers.
r=1n SrSr=r=1n rr+122rr+12= 12 r=1n r r+1              = 12 r=1n r2 + r=1n r               = 12 nn+1 2n + 16 + nn+12               = 12 nn+12 2n+13+1               = 14 n n+1 2n+1+33               = 14 n n+1 2n+43               = nn+1 2n+212
Hence,  r=1n SrSn= nn+1 n+26
Hence, the correct answer is option A.



Page No 21.20:

Question 1:

If S1 and S2 denote respectively the sum of first 100 natural numbers and the sum of their cubes, then the relation between S1 and S2 is __________.

Answer:

Let
S1 : the sum of first 100 natural numbers 
S2 : the sum of their cubes. 

S1= nn+12 = 100100+12S1= 5050S2= nn+122     = 100100+122     = 50502 S2= S12

Page No 21.20:

Question 2:

Let Sn and S'4 denote respectively the sum and the sum of the squares of first n natural numbers. If an=Sn'Sn, nN. Then a1, a2, a3,_____, an, _____ forms an ______ with ______ .

Answer:

Sn : The sum of first n natural numbers .
Sn : Sum of squares of first n natural numbers. 
If anSn'Sn ; nN.
a1= 11 = 1a2=12+221+2=53

a3=12+22+321+2+3=146=73 .......an=12+22 +.... +n21+2 +.... n =nn+1 2n+16nn+12= 2n+16×2an= Sn'Sn=2n+13 
Hence, common difference is
a2 − a153-1=23
aa73-53=23
Hence, common difference of  A.? is 23.

Page No 21.20:

Question 3:

The sum of first 25 odd natural numbers is _________.

Answer:

Sum of for 25 odd numbers :- 
Let S denote the sum of  first 25 odd numbers 
Here number are 1, 3, 5, 7, 9 --------- 49.
Here  first term is 1 

last term is 49
Common difference d is 2 
n = 25
 i.e Sum=n2 a+Tn= 252 1+49=25×502= 25×25
i.e sum of first 25 odd numbers = 625.

Page No 21.20:

Question 4:

The value of r=1n2r-1+12r is _______________.

Answer:

r=1n 2r - 1+12rFor r=1n 2r-1 = 1+3+5+7+ .......+ 2n+1
Here  first term is 1 

last term is 2n − 1
i.e sum is n2 (first term + last term) 
i.e n2 1+2n-1  =n2 2n
i.e r=1n 2r-1=n2       ...1 also r=1n 12r =12+122+123 + .... + 12nfirst term a=12, r = common ratio is 12Sum Sn= a1-rn1-r              = 121-12n1-12r=1n 12r=Sum=1-12n r=1n 2r-1+12r = n2+1-12n    r=1n 2r-1+12r =1+n2-12n

Page No 21.20:

Question 5:

The sum of n terms of the series 22 + 42 + 62 +......, is _______________.

Answer:

Sum of n terms of series  22 + 42 + 62 + .........
Let nth term of series be denoted byn
i.e n = (2n)2
 then r=1n Tr=r=1n 2r2= 4 r=1n r2= 4 nn+1 2n+16
i.e sum of n terms of 22 + 42 + 62 = 23 nn+1 2n+1

Page No 21.20:

Question 6:

14 + 24 + 34 +______+ n4 = _______________.

Answer:

Ans

Page No 21.20:

Question 7:

If S2 and S4 denote respectively the sum of the squares and the sum of the fourth powers of first n natural numbers, then S4S2=_______________.

Answer:

S2 : Sum of the squares of first n natural numbers.
S: Sum of the fourth powers of first n natural numbers.
To find :- S4S2
Since S4=n n+1 2n+1 3n2+3n-130S2=n n+1 2n+16Hence , S4S2=n n+1 2n+1 3n2+3n-130×n n+1 2n+1 ×6Hence ,  S4S2=3n2+3n - 15

Page No 21.20:

Question 8:

The value of 13+23+33+______+1031+2+3+______+10 ______________.

Answer:

13 + 23 + 33 + ...+ 1031 + 2 + ... + 10
Since 13+23+ ... +n3=nn+12213+23 + ... +103=1010+122= 10×1122= 552
also 1+2+ ... +n = nn+121+2 + ... +10 = 10112=5×112= 5513+23+ ... +103 1+2+ .. +n=55255=55

Page No 21.20:

Question 9:

If the sum of the squares of first n natural numbers exceeds their sum by 330, then n = _______________.

Answer:

Sum of squares of first n natural numbers is  n n+1 2n + 16
Sum of first n natural number is n n+12
According to given condition, Σn2 -Σn = 330
 i.e nn+1 2n+16 - nn+12 = 330i.e  nn+12 2n+13 -1 =330i.e  nn+12 2n+1- 33 = 330 i.e  nn+12 2n-23 = 330i.e  nn+1 2n-12 × 3 = 330i.e nn2-1=330×3i.e  n3-n= 990i.e  n3-n-990 = 0
i.e n3 − 10n2 + 10n2 − n − 990 = 0 − 99n + 99n
i.e n(n − 10) + 10n2 − 100n − 990 + 99n = 0
i.e n(n − 10) + 10n (n − 10) + 99(n−10) = 0
i.e (n − 10) (n2 + 10+ 99) = 0 
n = 10

Page No 21.20:

Question 10:

The sum of n terms of the series 312+512+22+712+22+32+_________ is _______________.

Answer:

Sum of n term of 312+512+22+712+22+32 
here nth term tn for above series is
 i.e Tn=2n+112+22 + .... +n2Tn=2n+1nn+12n+16=2n+1n+1 2n + 16Tn = 6nn+1i.e Tr= 6rr+1
Hence r=1n Tr= r=1n 6rr+1 = 6 r=1n 1rr+1=r=1n6  1r - 1r+1= 6 11 - 12 + 12 - 13 + 13 - 14 +....+ 1n - 1n+1  = 6 1 - 1n + 1 =6 nn+1
Hence, Sn sum of n term = 6nn+1

Page No 21.20:

Question 1:

Write the sum of the series 2 + 4 + 6 + 8 + ... + 2n.

Answer:

Sn=2+4+6+8+...+2n 
    =n24+n-12    =n22+2n    =nn+1

Page No 21.20:

Question 2:

Write the sum of the series 12 − 22 + 32 − 42 + 52 − 62 + ... + (2n − 1)2 − (2n)2.

Answer:

The given series can be rewritten as:

3-1 + 7-1 + 11-1 +...+4n-1 -1

=-3+7+11+...+4n-1=-n23×2+n-14=-n24n+2=-n2n+1

Page No 21.20:

Question 3:

Write the sum to n terms of a series whose rth term is r + 2r.

Answer:

Series whose rth term is r + 2r:

1+21+2+22+3+23+4+24+...+n+2n

Thus, we have:

Sn= 1+21+2+22+3+23+4+24+...+n+2n     = 1+2+3+4+...+n+2+22+23+24+...+2n     = nn+12+22n-12-1     = nn+12+2n+1-2

Page No 21.20:

Question 4:

If r=1nr=55, find r=1nr3.

Answer:

r=1nr3 = 13+23+33+...+n3          =nn+122          =r=1nr2          =552          =3025

Page No 21.20:

Question 5:

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then write the value of k.

Answer:

According to the question,
2+4+...+2n = k1+3+5+7+...+2n-1

2×nn+12= kn22×1+n-1×22nn+12= kn22+2n-2nn+1= kn22nn2+n= kn2k=n+1n



Page No 21.21:

Question 6:

Write the sum of 20 terms of the series 1+12(1+2)+13(1+2+3)+....

Answer:

Let the nth term be an.
Here, 
an=1n1+2+3+...+n=n+12
We know:
Sn=k=1nak
Thus, we have:
S20=k=120ak
     =12k=120k+1     =12k=120k+20     =1220212+20     =12230     =115

Page No 21.21:

Question 7:

Write the 50th term of the series 2 + 3 + 6 + 11 + 18 + ...

Answer:

We have,a1=2,a2=3=2+1,a3=6=2+1+3,a4=11=2+1+3+5,...a50=2+1+3+5+...50 terms=2+4922×1+49-1×2             As, the terms apart 2 are in A.P. with a=1 and d=2=2+4922+48×2=2+492×98=2+492=2+2401=2403

So, the 50th term of the given series is 2403.

Page No 21.21:

Question 8:

Let Sn denote the sum of the cubes of first n natural numbers and sn denote the sum of first n natural numbers. Then, write the value of
r=1nSrsr.

Answer:

We know that, Sr=13+23+33+...+r3=rr+122And, sr=1+2+3+...+r=rr+12As, Srsr=rr+122rr+12=rr+12=12r2+rNow,r=1nSrsr=r=1n12r2+r=12r=1nr2+r=1nr=12nn+12n+16+nn+12=12×nn+12×2n+13+1=nn+142n+1+33=nn+142n+43=nn+14×2n+23=nn+1n+26



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