NIMCET 2011 Question Paper with solutions

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Question : 27 of 120

Marks: +1, -0

If for

n ∈ N, ∑↙{k=0}↖{2 n}(-1)^{k}[(\table 2 n ; k)]^{2}=A

, then the value of

∑(-1)^{k}(k-2 n)[(\table 2 n ; k)]^{2}

is

nA
-nA
0
A

Solution:

We know,

(1+x)^{2 n}=C_{0}+C_{1} x+. . .+C_{2 n} x^{2 n}

=C_{0} x^{2 n}+C_{1} x^{2 n-1}+. . .+C_{2 n}

...(i)
and

(1-x)^{2 n}=C_{0}-C_{1} x+. . .+C_{2 n} x^{2 n}

......(ii)

On multiplying Eqs. (i) and (ii), we get

(1-x^{2})^{2 n}=(C_{0} x^{7 n}+C_{1} x^{2 n-1}+. . .+C_{2 b})

x(C_{0}-C_{1} x+. . .+C_{2 n} x^{2 n})

On equating coefficient of

x^{2 n}

on both sides, we get

(-1)^{n} C_{n}=C_{n}^{2}-C_{1}^{2}+C_{2}^{2}-. . .+C_{20}^{2}

=∑↙{i=0}↖{2 n}(-1)^{k}(\table 2 n ; k)^{2}=A \;\;

(given)

. . .(i)

Similarly we can find

C_{1}^{2}-2 C_{2}^{2}+3 C_{3}^{2}-. . .-2 n C_{2 n}^{2}

=∑↙{k=0}↖{2 n}(-1)^{k}(\table 2 n ; k)^{2}=\(-1)^{n} ⋅ n ⋅ C_{n}

∴ \;\; ∑↙{k=0}↖{2 n}(-1)^{k} k(\table 2 n ; k)^{2}-2 n ∑↙{k=0}↖{2 n}(-1)^{k}(\table 2 n ; k)^{2}

=(-1)^{n} ⋅ n ⋅ C_{n}-2 n(-1)^{n} ⋅ C_{n}

=-(-1)^{n} n ⋅ C_{n}=-n A \;\;

[from Eq. (i)]

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