If Sn=∑r=1n1+2+22+ r terms 2r,then Snis equal to(A)2n-(n+1)(B)n×(n+1)/2(C)(n2+3n+2)/6(D)n-1+(1/2n) - Maths - Sequences and Series

Question

I f   S n = ∑ r = 1 n 1 + 2 + 2 2 + r   t e r m s
  2 r , t h e n   S n i s   e q u a l   t o ( A
) 2 n - ( n + 1 ) ( B ) n × ( n + 1 ) / 2 ( C ) ( n 2 + 3 n +
2 ) / 6 ( D ) n - 1 + ( 1 / 2 n )

Punit Sharma · Accepted Answer

Dear Student,

As,

Sn=∑r=1n1+2+22 r terms2r (i)since, 1+2+22
r terms is a geometric progression,and sum of r terms of GP = A×xr-1x-1  where x is the common difference and A is the first termhence, 1+2+22
r terms = 1×2r-12-1Replacing this value in equation (i), Sn=∑r=1n1×2r-12-1×12ror, Sn=∑r=1n2r-12r=∑r=1n2r2r-12ror, Sn=∑r=1n1-∑r=1n12r=n-121-12n1-12                                        as ∑r=1n1=n and ∑r=1n12r will be a GP whose sum = A×xr-1x-1 or, Sn=n-121-12n12=n-1+12n

Thus, Sn=n-1+12n

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I f S n = ∑ r = 1 n 1 + 2 + 2 2 + . . . . . . . r t e r m s 2 r , t h e n S n i s e q u a l t o ( A ) 2 n - ( n + 1 ) ( B ) n × ( n + 1 ) / 2 ( C ) ( n 2 + 3 n + 2 ) / 6 ( D ) n - 1 + ( 1 / 2 n )

$I f S_{n} = \sum_{r = 1}^{n} \frac{1 + 2 + 2^{2} + . . . . . . . r t e r m s}{2^{r}}, t h e n S_{n} i s e q u a l t o (A) 2^{n} - (n + 1) (B) n \times (n + 1) / 2 (C) (n^{2} + 3 n + 2) / 6 (D) n - 1 + (1 / 2^{n})$