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Blazing goIITian

total posts: 1251
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here is my shot at the answer -->

taking

(i.j) (ⁿC_i . ⁿC_j) = x

^0<i<j<=n

differentiating with respect to x

n(1 + x)

^(n-1)

^n-1 )

on putting x =1

n(2)

^n-1

or on squaring

n^2*2

^{2n - 2} =
(C_1^2 + 2^2C_2^2 + ...........n ^2C_n^2 ) +

now

we have to find the value of

(C_1^2 + 2^2C_2^2 + ...........n ^2C_n^2 )

for that we proceed as follows --->

n(1 + x)

^(n-1)

^n-1 ) ---1

replacing x by 1/x

n(1 + 1/x)

^n-1=

^n-1 ) ---2

1 * 2

[n^2(1 + x)

^2n-2

^n-1

^n-1 ) *

^n-1 )

constant term in lhs is

n^2

* coefficient of x^n-1 in (1 + x)^(2n-2)]

= n^2

*^{2n - 2}

_n-1

constant term in rhs is -->

(C_1^2 + 2^2C_2^2 + ...........n ^2C_n^2 )

hence

^{2n - 2}

_n-1

thus

n^2

^{(2n - 2)} = n^2

^{2n - 2}

_n-1

^{(2n - 3)} - n^2

^{2n - 2}

_n-1

]

Impossible To be Impossible is Impossible

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