Q no 7 â ‹Q7 If Sn = =∑r =1n1+2+22 r terms2r, then Sn is equal - Maths - Sequences and Series

Question

Q no 7

â€‹Q7 If Sn = = ∑ r   = 1 n 1
+ 2 + 2 2 r   terms 2 r , then Sn is equal to

(A) 2n – (n +1) (B) n × (n +1) /2

(C) (n2 + 3n +2 )/6 (D) n – 1 + (1/2n)

Punit Sharma · Accepted Answer

Dear Student,

The solution is as under:

Sn=∑r=1n1+2+22+ r terms2rsince, 1+2+22+
r terms is GP with common difference as 2Hence sum=Arn-1r-1=12r-12-1=2r-1or, Sn=∑r=1n1+2+22+
r terms2r=∑r=1n2r-12ror, Sn=∑r=1n2r2r-∑r=1n12ror, Sn=∑r=1n1-∑r=1n12ror, Sn=n-121-12n1-12       as ∑r=1n1=n and ∑r=1n12r will be GP with common ratio as 12or,  Sn=n-121-12n12 Sn=n-1+12n

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Q no 7 ​Q7. If Sn = = ∑ r = 1 n 1 + 2 + 2 2 . . . r terms 2 r , then Sn is equal to (A) 2n – (n +1) (B) n × (n +1) /2 (C) (n2 + 3n +2 )/6 (D) n – 1 + (1/2n)

Q no 7

Q7. If S_n = $= \sum_{r = 1}^{n} \frac{1 + 2 + 2^{2} . . . r terms}{2^{r}}$ , then Sn is equal to

(A) 2ⁿ – (n +1) (B) n × (n +1) /2

(C) (n² + 3n +2 )/6 (D) n – 1 + (1/2ⁿ)