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If sum(r=0)^(10)((r+2)/(r+1))^^n Cr=(2^8...

If `sum_(r=0)^(10)((r+2)/(r+1))^^n C_r=(2^8-1)/6` , then `n` is `8` b. `4` c. `6` d. `5`

A

8

B

4

C

6

D

5

Text Solution

Verified by Experts

The correct Answer is:
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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CENGAGE-BINOMIAL THEOREM-Exercise (Single)
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  3. If sum(r=0)^(10)((r+2)/(r+1))^^n Cr=(2^8-1)/6 , then n is 8 b. 4 c. 6 ...

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  12. The value of sum(r=0)^(50)(-1)^r(^(50)Cr)/(r+2) is equal to 1/(50xx51)...

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