coefficient of x^5 in expansion (1+x^2 )^5 (1+x)^4 equal to (a) 40 (b) 45 (c) 50 (d) 55 (e) 60


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Question Number 103513 by bemath last updated on 15/Jul/20

$c o e f f i c i e n t o f x^{5} i n e x p a n s i o n$ ${(1 + x^{2})}^{5} {(1 + x)}^{4} e q u a l t o$ $(a) 40 (b) 45 (c) 50 (d) 55$ $(e) 60$
	Answered by OlafThorendsen last updated on 15/Jul/20

	$C_{5}^{1} C_{4}^{3} + C_{5}^{2} C_{4}^{1} = 5 \times 4 + \frac{5!}{3! 2!} \times 4 = 60$ $\Rightarrow (e)$
	Commented by bemath last updated on 15/Jul/20

	$h o w d o y o u g o t i t ?$
	Commented by OlafThorendsen last updated on 15/Jul/20

	${(1 + x^{2})}^{5} = \underset{k = 0}{\sum^{5}} C_{5}^{k} x^{2 k}$ ${(1 + x)}^{4} = \underset{k = 0}{\sum^{4}} C_{4}^{k} x^{k}$ $To obtain x^{5} two solutions only :$ $x^{2} \times x^{3} or x^{4} \times x^{1}$ $Then C_{5}^{2} C_{4}^{3} + C_{5}^{4} C_{4}^{1}$
	Answered by Rio Michael last updated on 15/Jul/20

	${(1 + x^{2})}^{5} = \underset{r = 0}{\sum^{5}}^{5} C_{r} x^{2 r} . . . . . . (i)$ ${(1 + x)}^{4} = \underset{r = 0}{\sum^{4}}^{4} C_{r} x^{r} . . . . . . (i i)$ $\Rightarrow {(1 + x^{2})}^{5} {(1 + x)}^{4} = \underset{r = 0}{\sum^{5}}^{5} C_{r} x^{2 r} . \underset{r = 0}{\sum^{4}}^{4} C_{r} x^{r}$ $when r = 1 for (i) and r = 3 for (i i) we get x^{5}$ $\Rightarrow possibility is :^{5} C_{1} .^{4} C_{3}$ $also when r = 2 for (i) and r = 1 for (i i) we get x^{5}$ $\Rightarrow possibility is :^{5} C_{2} .^{4} C_{1}$ $\Rightarrow all possibilities are :^{5} C_{1} .^{4} C_{3} +^{5} C_{2} .^{4} C_{1}$ $an extended way of the solution above .$
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