MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Summation by Sigma Method

Prove that ...

Question

Prove that $n \sum k = 1 k 2^{- k} = 2 [1 - 2^{- n} - n {.2}^{- (n + 1)}]$ .

Open in App

Solution

$S_{n} = {1.2}^{- 1} + {2.2}^{- 2} + {3.2}^{- 3} + . . . . . . . . + n {.2}^{- n}$
$2^{- 1} S_{n} = 2^{- 2} + {2.2}^{- 3} + . . . . . . . . + (n - 1) {.2}^{- n} + n {.2}^{- (n + 1)}$
$(1 - 2^{- 1}) S_{n} = 2^{- 1} + 2^{- 2} + 2^{- 3} + . . . . . . + 2^{- n} - n {.2}^{- (n + 1)}$
$2^{- 1} S_{n} = 2^{- 1} \frac{2^{- n} - 1}{2^{- 1} - 1} - n {.2}^{- (n + 1)}$
$S_{n} = 2 (1 - 2^{- n} - n {.2}^{- (n + 1)})$

Suggest Corrections

thumbs-up

0

Similar questions

K = 1^{\circ}

, then prove that

88 \sum n = 0 \frac{1}{cos n K . cos (n + 1) K} = \frac{cos K}{{sin}^{2} K}

Prove that : $\frac{(2 n + 1)!}{n!} = 2 n 1.3.5.... (2 n - 1) (2 n + 1)$

\frac{2^{n} + 2^{n - 1}}{2^{n + 1} - 2^{n}} = \frac{3}{2}

Q. Prove by mathematical induction that
n. 1 + ( n - 1) 2 + (n - 2) 3 + ..... + 2 (n -1) + 1.n =

\frac{n}{6}

\frac{(2 n + 1)!}{n!}

= 2ⁿ [1 · 3 · 5 ... (2n − 1) (2n + 1)]

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Summation by Sigma Method

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Summation by Sigma Method

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon