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if (1+a)^(n)=.^(n)C(0)+.^(n)C(1)a++.^(n)...

if `(1+a)^(n)=.^(n)C_(0)+.^(n)C_(1)a++.^(n)C_(2)a^(2)+ . . .+.^(n)C_(n)a^(n)`, then prove that
`(.^(n)C_(1))/(.^(n)C_(0))+(2(.^(n)C_(2)))/(.^(n)C_(1))+(3(.^(n)C_(3)))/(.^(n)C_(2))+. . . +(n(.^(n)C_(n)))/(.^(n)C_(n-1))=` Sum of first n natural numbers.

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The correct Answer is:
`L.H.S.=(.^(n)C_(1))/(.^(n)C_(0))+(2.^(n)C_(2))/(.^(n)C_(1))+(3.^(n)C_(3))/(.^(n)C_(2))+ . . .+(n.^(n)C_(n))/(.^(n)C_(n-1))`
`=(n)/(1)+(2.n(n-1))/((2)/(n))+(3.n(n-1)(n-2))/((3.2.1)/((n(n-1))/(2))+ . . .+(n.1)/(n))`
`=n+(n-1)+(n-2)+ . . .+1`
`=1+2+3+ . . .+n`
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If (1+a)^(n)=.^(n)C_(0)+.^(n)C_(1)a+.^(n)C_(2)a^(2)+ . . +.^(n)C_(n)a^(n) , then prove that .^(n)C_(1)+2.^(n)C_(2)+3.^(n)3C_(3)+ . . .+n.^(n)C_(n)=n.2^(n-1) .

(.^(n)C_(1))/(2)-(2(.^(n)C_(2)))/(3)+(3(.^(n)C_(3)))/(4)-....+(-1)^(n+1)(n(.^(n)C_(n)))/(n+1)=

(*^(n)c_(0))^(2)+3(.^(n)C_(1))^(2)+5(.^(n)C_(2))^(2)+ ......+(2n+1)(.^(n)C_(n))^(2)

Find .^(n)C_(1)-(1)/(2).^(n)C_(2)+(1)/(3).^(n)C_(3)- . . . +(-1)^(n-1)(1)/(n).^(n)C_(n)

Prove that (.^(n)C_(0))/(1)+(.^(n)C_(2))/(3)+(.^(n)C_(4))/(5)+(.^(n)C_(6))/(7)+"....."+= (2^(n))/(n+1)

The value of .^(n)C_(0).^(n)C_(n)+.^(n)C_(1).^(n)C_(n-1)+...+.^(n)C_(n).^(n)C_(0) is

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