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If _{r = 0}^{n}( {r + 2}{r + 1});;; ^{n}C_{r} = {2^{8} - 1}{

Question

If $r = 0 \sum n (\frac{r + 2}{r + 1})^{n} C_{r} = \frac{2 ^{8} - 1}{6}$ , then $n =$ .

A

8

B

4

C

6

D

5

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Sophia Discussed

If

r = 0 \sum n (\frac{r + 2}{r + 1})^{n} C_{r} = \frac{2 ^{8} - 1}{6}

, then

n =

.

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Text solutionVerified

r = 0 \sum n \frac{r + 2}{r + 1}^{n} C_{r} = \frac{2 ^{8} - 1}{6}

\Rightarrow \displaystyle \sum_{r = 0}^{n} \left [1 + \dfrac {1}{r + 1}\right ]\ ^{n}C_{r} = \dfrac {2^{8} - 1}{6}

\Rightarrow 2^{n} + r = 0 \sum n \frac{1}{n + 1} .^{n + 1} C_{r + 1} = \frac{2 ^{8} - 1}{6}

\Rightarrow 2^{n} + \frac{2 ^{n + 1} - 1}{n + 1} = \frac{2 ^{8} - 1}{6}

\Rightarrow \frac{2 ^{n} ( n + 3 ) - 1}{n + 1} = \frac{2 ^{8} - 1}{6}

\Rightarrow \frac{2 ^{n} ( n + 1 + 2 ) - 1}{n + 1} = \frac{2 ^{5} ( 6 + 2 ) - 1}{6}

On comparing, we get

n + 1 = 6 \Rightarrow n = 5

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Question Text	If $r = 0 \sum n (\frac{r + 2}{r + 1})^{n} C_{r} = \frac{2 ^{8} - 1}{6}$ , then $n =$ .
Answer Type	Text solution:1
Upvotes	150