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If `n` is a positive integer, prove that `1-2n+(2n(2n-1))/(2!)-(2n(2n-1)(2n-2))/(3!)++(-1)^(n-1)(2n(2n-1)(n+2))/((n-1)!)= (-1)^(n+1)(2n)!//2(n !)^2dot`

Text Solution

Verified by Experts

Given sum is
`S = .^(2n)C_(0) - .^(2n)C_(1) + .^(2n)C_(2)- "……" + (-1)^(n-1) xx .^(2n)C_(n-1)`
` :. 2S = 2(.^(2n)C_(0) - .^(2n)C_(1) + .^(2n)C_(2) -"….." + (-1)^(n-1) xx .^(2n)C_(n-1))`

`= (.^(2n)C_(0) + .^(2n)C_(2n)) - (.^(2n)C_(1) + .^(2n)C_(2n-1)) + (.^(2n)C_(2) + .^(2n)C_(2n-2))`
`-"....."+((-1)^(n-1) xx .^(2n)C_(n-1) + (-1)^(n+1) xx .^(2n)C_(n+1)))`
` = [.^(2n)C_(0) - .^(2n)C_(1) + .^(2n)C_(2) - .^(2n)C_(3) + "......" + (-1)^(n-1) xx .^(2n)C_(n+1)`
`+ (-1)^(n) xx .^(2n)C_(n) + (-1)^(n+1) xx .^(2n)C_(n+1) + "......" + .^(2n)C_(2n))]`
`-(-1)^(n) xx .^(2n)C_(n)`
` = (1-1)^(2n)+ (-1)^(n+1)xx.^(2n)C_(n)`
`:. S = (-1)^(n+1) ((2n)!)/(2(n!)^(2))`
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