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Prove that .^(1)P(1)+2*.^(2)P(2)+3*.^(...

Prove that
`.^(1)P_(1)+2*.^(2)P_(2)+3*.^(3)P_(3)+ . . .+n*.^(n)P_(n)=.^(n+1)P_(n+1)-1`.

Text Solution

Verified by Experts

The correct Answer is:
`x=n+3`
We have, `.^(n)C_(n-r)+3^(n)C_(n-r+1)+3^(n)C_(n-r+2)+.^(n)C_(n-r+3)=.^(n)C_(r)`
`hArr (.^(n)C_(n-r)+.^(n)C_(n-r+1))+2(.^(n)C_(n-r+1)+.^(n)C_(n-r+2))+(.^(n)C_(n-r+2)+.^(n)C_(n-r+3))=.^(x)C_(r)`
`hArr.^(n+1)C_(n-r+1)+2^(2N=1)C_(n-r+2)+.^(n+1)C_(n-r+3)=.^(x)C_(r)`
`hArr(.^(n+1)C_(n-r+1)+.^(n+1)C_(n-r+2))+(.^(n+1)C_(n-r+2)+.^(n+1)C_(n-r+3))=.^(x)C_(r)`
`hArr.^(n+2)C_(n-r+2)+.^(n+2)C_(n-r+3)=.^(x)C_(r)`
`hArr .^(n+3)C_(n-r+3)=.^(x)C_(r)`
`hArr .^(n+r)C_(r)+.^(x)C_(r)" "[because .^(n)C_(r)=.^(n)C_(n-r)]`
Hence, `x=n+3`
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