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If `sum_(r=0)^(n) ((r+2)/(r+1))""^(n)C_(r)=(2^(8)-1)/(6)`. then n is
A. 8
B. 4
C. 6
D. 5

1 Answer

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Best answer
Correct Answer - D
`underset(r=0)overset(n)sum((r+1)/(r+1)+(1)/(r+1)).^(n)C_(1)`
`= underset(r=0)overset(n)sum.^(n)C_(r)+underset(r=0)overset(n)sum(.^(n)C_(r))/(r+1)`
`= 2^(n)+(1)/(n+1).underset(r=0)overset(n)sum.^(n+1)C_(r+1)`
`= 2^(n)+(2^(n+1)-1)/(n+1)`
`= ((n+3)2^(n)-1)/(n+1)`
`= (2^(8)-1)/(6)`
`rArr n = 5`

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