RD Sharma XI 2020 Volume 1 Solutions for Class 12 Commerce Math 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 Math 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 Math 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 12 Commerce Math are prepared by experts and are 100% accurate.
Page No 21.10:
Question 1:
13 + 33 + 53 + 73 + ...
Answer:
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
Page No 21.10:
Question 2:
22 + 42 + 62 + 82 + ...
Answer:
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
Page No 21.10:
Question 3:
1.2.5 + 2.3.6 + 3.4.7 + ...
Answer:
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
Page No 21.10:
Question 4:
1.2.4 + 2.3.7 +3.4.10 + ...
Answer:
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
Page No 21.10:
Question 5:
1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...
Answer:
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
Page No 21.10:
Question 6:
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...
Answer:
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
Page No 21.10:
Question 7:
3 × 12 + 5 ×22 + 7 × 32 + ...
Answer:
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
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 be the nth term of the given series.
Thus, we have:
(i)
Let be the sum of n terms of the given series.
Now,
(ii)
Let be the sum of n terms of the given series.
Now,
(iii)
Let be the sum of n terms of the given series.
Now,
(iv)
Let be the sum of n terms of the given series.
Now,
(v)
Let be the sum of n terms of the given series.
Now,
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 be the nth term of the given series.
Thus, we have:
For n = 20, we have:
Therefore, the 20th term of the given series is 1680.
Now, let be the sum of n terms of the given series.
Thus, we have:
For n = 20, we have:
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 be the nth term and be the sum to n terms of the given series.
Thus, we have:
...(1)
Equation (1) can be rewritten as:
...(2)
On subtracting (2) from (1), we get:
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:
Now,
Page No 21.18:
Question 2:
2 + 5 + 10 + 17 + 26 + ...
Answer:
Let be the nth term and be the sum of n terms of the given series.
Thus, we have:
...(1)
Equation (1) can be rewritten as:
...(2)
On subtracting (2) from (1), we get:
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:
Now,
Page No 21.18:
Question 3:
1 + 3 + 7 + 13 + 21 + ...
Answer:
Let be the nth term and be the sum of n terms of the given series.
Thus, we have:
...(1)
Equation (1) can be rewritten as:
...(2)
On subtracting (2) from (1), we get:
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:
Now,
Page No 21.18:
Question 4:
3 + 7 + 14 + 24 + 37 + ...
Answer:
Let be the nth term and be the sum of n terms of the given series.
Thus, we have:
...(1)
Equation (1) can be rewritten as:
...(2)
On subtracting (2) from (1), we get:
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:
Now,
Page No 21.18:
Question 5:
1 + 3 + 6 + 10 + 15 + ...
Answer:
Let be the nth term and be the sum of n terms of the given series.
Thus, we have:
....(1)
Equation (1) can be rewritten as:
...(2)
On subtracting (2) from (1), we get:
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:
Now,
Page No 21.18:
Question 6:
1 + 4 + 13 + 40 + 121 + ...
Answer:
Let be the nth term and be the sum to n terms of the given series.
Thus, we have:
...(1)
Equation (1) can be rewritten as:
...(2)
On subtracting (2) from (1), we get:
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:
Page No 21.18:
Question 7:
4 + 6 + 9 + 13 + 18 + ...
Answer:
Let be the nth term and be the sum of n terms of the given series.
Thus, we have:
...(1)
Equation (1) can be rewritten as:
...(2)
On subtracting (2) from (1), we get:
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,
Page No 21.18:
Question 8:
2 + 4 + 7 + 11 + 16 + ...
Answer:
Let be the sum of n terms and be the nth term of the given series.
Thus, we have:
...(1)
Equation (1) can be rewritten as:
...(2)
On subtracting (2) from (1), we get:
Page No 21.18:
Question 9:
Answer:
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
Page No 21.18:
Question 10:
Answer:
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
Page No 21.18:
Question 1:
The sum to n terms of the series is
(a)
(b)
(c)
(d)
Answer:
(d)
Let be the nth term of the given series.
Thus, we have:
Now,
Let be the sum of n terms of the given series.
Thus, we have:
Page No 21.19:
Question 2:
The sum of the series is
(a)
(b)
(c)
(d) none of these
Answer:
(c)
Let
Page No 21.19:
Question 3:
The value of is equal to
(a)
(b)
(c)
(d) none of these
Answer:
(b)
We have:
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
Now,
Page No 21.19:
Question 5:
If Sn = , then Sn is equal to
(a) 2n − n − 1
(b)
(c)
(d) 2n − 1
Answer:
(c)
We have:
Sn =
Page No 21.19:
Question 6:
If to n terms is S, then S is equal to
(a)
(b)
(c)
(d) n2
Answer:
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
Page No 21.19:
Question 7:
Sum of n terms of the series .... is
(a)
(b) 2n (n + 1)
(c)
(d) 1
Answer:
(c)
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
Page No 21.19:
Question 8:
The sum of 10 terms of the series .... is
(a)
(b)
(c)
(d)
Answer:
(a)
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of 10 terms of the given series.
Thus, we have:
Page No 21.19:
Question 9:
The sum of the series 12 + 32 + 52 + ... to n terms is
(a)
(b)
(c)
(d)
Answer:
(b)
Let be the nth term of the given series.
Thus, we have:
Now, let be the sum of n terms of the given series.
Thus, we have:
Page No 21.19:
Question 10:
The sum of the series ..... to n terms is
(a)
(b)
(c)
(d)
Answer:
(b)
Let be the nth term of the given series.
Thus, we have:
Now,
Let be the sum of n terms of the given series.
Thus, we have:
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 equals
(a)
(b)
(c)
(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
; formula for sum of cubes.
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.
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 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 an = ; nâN.
Hence, common difference is
a2 − a1 =
a3 − a2 =
Hence, common difference of A.? is .
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
n = 25
i.e sum of first 25 odd numbers = 625.
Page No 21.20:
Question 4:
The value of is _______________.
Answer:
Here first term is 1
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 byâ¤n
i.e â¤n = (2n)2
i.e sum of n terms of 22 + 42 + 62 =
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 _______________.
Answer:
S2 : Sum of the squares of first n natural numbers.
S4 : Sum of the fourth powers of first n natural numbers.
To find :-
Page No 21.20:
Question 8:
The value of ______________.
Answer:
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
Sum of first n natural number is
According to given condition,
i.e n3 − 10n2 + 10n2 − n − 990 = 0 − 99n + 99n
i.e n2 (n − 10) + 10n2 − 100n − 990 + 99n = 0
i.e n2 (n − 10) + 10n (n − 10) + 99(n−10) = 0
i.e (n − 10) (n2 + 10n + 99) = 0
⇒ n = 10
Page No 21.20:
Question 10:
The sum of n terms of the series _________ is _______________.
Answer:
Sum of n term of
here nth term tn for above series is
Page No 21.20:
Question 1:
Write the sum of the series 2 + 4 + 6 + 8 + ... + 2n.
Answer:
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:
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:
Thus, we have:
Page No 21.20:
Question 4:
If .
Answer:
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,
Page No 21.21:
Question 6:
Write the sum of 20 terms of the series
Answer:
Let the nth term be .
Here,
We know:
Thus, we have:
Page No 21.21:
Question 7:
Write the 50th term of the series 2 + 3 + 6 + 11 + 18 + ...
Answer:
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
.
Answer:
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