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The value of .^(n)C(1)+.^(n+1)C(2)+.^(n+...

The value of `.^(n)C_(1)+.^(n+1)C_(2)+.^(n+2)C_(3)+"….."+.^(n+m-1)C_(m)` is equal to a. `.^(m+n)C_(n) - 1` b. `.^(m+n)C_(n-1)` c. `.^(m)C_(1) + ^(m+1)C_(2) + ^(m+2)C_(3) + "…." + ^(m+n-1)C_(n)` d. `.^(m+n)C_(m) - 1`

A

`.^(m+n)C_(n) - 1`

B

`.^(m+n)C_(n-1)`

C

`.^(m)C_(1) + .^(m+1)C_(2) + .^(m+2)C_(3) + "…." + .^(m+n-1)C_(n)`

D

`.^(m+n)C_(m) - 1`

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
`.^(n)C_(1)+.^(n+1)C_(2)+.^(n+2)C_(3)+"……"+.^(n+m-1)C_(m)`
`= .^(n)C_(n-1)+.^(n+1)C_(n-1)+.^(n+2)C_(n-1)+"……"+.^(n+m-1)C_(n-1)`
`=` Coefficient of `x^(n-1)` in `(1+x)^(n) [((1+x)^(m) -1)/((1+x) - 1)]`
= Coefficient of `x^(n-1)` in `((1+x)^(m+n) -(1+x)^(n))/(x)`
`=` Coefficient of `x^(n)` in `[(1+x)^(m+n) - (1+x)^(n)]`
`= .^(m+n)C_(n) - 1`
Similarly, we can prove
`.^(m)C_(1)+.^(m+1)C_(2)+.^(m+2)C_(3)+"...."+.^(m+n-1)C_(n)=.^(m+n)C_(m)-1`
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