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Prove that .^(n)C(0) +5 xx .^(n)C(1) + 9...

Prove that `.^(n)C_(0) +5 xx .^(n)C_(1) + 9 xx .^(n)C_(2) + "…." + (4n+1) xx .^(n)C_(n) = (2m+1) 2^(n)`.

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`S = .^(n)C_(0) + 5 xx .^(n)C_(1)+9xx.^(n)C_(2)+"....."(4n-3)xx.^(n)C_(n-1)+(4n+1)xx.^(n)C_(n)"......"(1)`
`:. S = (4n+1).^(n)C_(n)+(4n-3).^(n)C_(n-1)+"...."+5.^(n)C_(1)+.^(n)C_(n)"....."(2)`
Adding (1) and (2), we get
`2S = (4n+2)(.^(n)C_(0)+.^(n)C_(1)+"....."+.^(n)C_(n-1)+.^(n)C_(n))`
`= (4n+2)2^(n)`
`rArr S = (2n+1)2^(n)`
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