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AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Section: Mathematics

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Question : 17 of 160

Marks: +1, -0

The coefficient of

x^5

in the expansion of

(x^2+2x+3)^5

is

1052
540
480
1020

Solution:

Consider the expression,

(3+2 x+x^{2} )^{5} = \frac{\sum n!}{(p!)(q!)(r!)}

(3)^{p}(2 x)^{q} (x^{2} )^{r}

Here,

p+q+r=n=5

For

x^{5}, q+2 r=5

This gives,

r=0, q=5, p=0

r=1, q=3, p=1

r=2, q=1, p=2

So, we get,

(3+2 x+x^{2} )^{5} = \left( \frac{5!}{0!5!0!} (3)^{0}(2)^{5}(1)^{0} \right)

+ \frac{5!}{1!3!1!} (3)^{1}(2)^{3}(1)^{1} + \frac{5!}{2!1!2!} (3)^{2}(2)^{1}(1)^{2})

=32+20(3)(8)+30(9)(2)

=1052

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