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sum(m=1)^(n)(sum(k=1)^(m)(sum(p=k)^(m)"^...

`sum_(m=1)^(n)(sum_(k=1)^(m)(sum_(p=k)^(m)"^(n)C_(m)*^(m)C_(p)*^(p)C_(k)))=`

A

`3^(n)-2^(n)`

B

`4^(n)-3^(n)`

C

`3^(n)+2^(n)`

D

`4^(n)-1`

Text Solution

Verified by Experts

The correct Answer is:
B
`(b)` `sum_(m=1)^(n)'^(n)C_(m)(sum_(k=1)^(m)(sum_(p=k)^(m)(m!)/(p!(m-p)!)*(p!)/(k!(p-k)!)))`
`=sum_(m=1)^(n)'^(n)C_(m)(sum_(k=1)^(m)(sum_(p=k)^(m)'^(m-k)C_(p-k))(m!)/(k!(m-k)!))`
`=sum_(m=1)^(n)'^(n)C_(m)(sum_(k=1)^(m)2^(m-k)*^(m)C_(k))`
`=sum_(m=1)^(n)'^(n)C_(m)((1+2)^(m)-2^(m))=sum_(m=1)^(n)('^(n)C_(m)3^(m)-^(n)C_(m)2^(m))`
`=(1+3)^(n)-1-(1+2)^(n)+1=4^(n)-3^(n)`
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