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UPSC NDA I 2014 Mathematics Paper

For the next three (03) items that follow:

In the expansion of

\left(x^3 - \frac{1}{x^2}\right)^n

where n is a positive integer,

the sum of the coefficients of

x^5

and

x^{10}

is 0

[NDA I 2014]

Question : 36 of 120

Marks: +1, -0

Solution:

T_{r+1} = \binom{n}{r} (x^3)^{n-r} \left(-\frac{1}{x^2}\right)^r = (-1)^r \binom{n}{r} x^{3n-5r}

3n-5r=5

⇒

r = \frac{3n-5}{5}

3n-5r=10

\Rightarrow r = \frac{3n-10}{5}

Sum of

x^5

and

x^{10}

coefficients

= (-1)^{\frac{3n-5}{5}} \;\; \binom{n}{\frac{3n-5}{5}} + (-1)^{\frac{3n-10}{5}} \;\; \binom{n}{\frac{3n-10}{5}} = 0

By Hit n Trial, n = 15 satisfies the equation.

The middle terms are 7th and 8th and same as coefficients of

x^5

and

x^{10}

Therefore,sum is zero

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