यदि `(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+.....+C_(n)x^(n)`. साबित कीजिए कि
`C_(0)""^(2)+C_(1)""^(2)+C_(2)""^(2)+......+C_(n)""^(2)=((2n)!)/(n!n!)=(1.3.5...(2n-1).2^(n))/(n!)`

Explanation

Given ` (1 +x)^(n) = C_(0) + C_(1)x + C_(2)x^(n-2) + …+ C_(n)` … (i)
Now , `(x + 1)^(n) = C_(0) x^(n) + C_(1) x^(n-1) + C_(2) x^(n-2) + …+ C_9n)` …(ii)
On multiply Eqs.(i) and (ii), we get
` (x +1)^(2n) = (C_(0)x^(n) + C_(1) x^(n) + C_(1) x^(n-1) + C_(2) x^(n-2) + ...+ C_(n))`...(iii)
Now , coefficient of ` x^(n)` in RHS
` = C_(0)^(2) + C_(1)^(2) + C_(2)^(2) + ...+ C_(n)^(2)`
And coefficient of ` x^(n)` in LHS `= ""^(2n)C_(n)= (2n!)/(n!n!)`
` = (1*2*3*4*5...* (2n-1) 2n)/(n!n!) = (1*2*3*...(2n-1)2^(n)n!)/(n!n!)`
But Eq.(iii) is an identity , therefore coefficient of `x^(r)` in RHS
= coefficient of ` x^(n)` in LHS ,
` rArr C_(0)^(2) + C_(1)^(2) + C_(2)^(2) + ...+ C_(n)^(2) = (2n!)/(n!n!) `
` = (1*3*5)*...*(2n-1))/(n!) *2^(n)`
Aliter
Given , ` (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + ... + C_(n)^(2)` ...(i)
Now , ` (1 + (1)/(x))^(n) = C_(0) + (C_(1))/(x) + C_(2))/(x^(2)) + ... + (C_(n) x^(n))`
` xx(C_(0) + (C_(1))/(x) + (C_(2))/(x^(2)) + ...+ (C_(n))/(x^(n)))`...(iii)
Now , constant term in RHS ` = C_(0)^(2) + C_(1)^(2) + C_(2)^(2) + ...+ C_(n)^(2) `
Constant term in LHS = Constant term in ` (1 + x^(2n))/(x^(n))`
= Coefficient of ` x^(n) " in" (1 + x)^(2n) = ""^(2n)C_(n) + (2n!)/(n!n!)`
`= (n!2^(2)[1*3*5*...(2n -1)])/(n!n!) = (2^(n) [1*3*5...(2n-1)])/(n!)`
But Eq. (iii) ia an identity , therefore the contant term in RHS = constant term in LHS .
`rArr C_(0)^(2) + C_(1) ^(2) + C_(2)^(2) + ...+ C_(n)^(2) = (2n!)/(n!n!) = ({1*3*5...(2n-1)})/(n!) 2^(n)` .