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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)`. |
Answer» `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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