NIMCET 2008 Question Paper with solutions

Question : 11 of 120

Marks: +1, -0

In the expression

(x+1)(x+4)(x+9)

(x+16) . . .(x+400)

the coefficient of

x^{19}

Solution:

(x+1)(x+4)(x+9) . . .(x+400)

=(x+1)(x+z^{2})(x+3^{2}) . . .(x+20^{2})

So, coefficient of

x^{19}

will be

1^{2}+z^{2}+. . .+(20)^{2} \;\; [∵ Σ n^{2}=\;{n(n+1)(2 n+1)}/{6}]

=\;{20 × 21 × 41}/{6}=41 × 70=2870

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