Find the coefficient of x^{5} in the product | vedclass

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(A) Using the Binomial Theorem,the expansions are:$(1+2x)^{6} = \sum_{k=0}^{6} \binom{6}{k} (2x)^k = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^

Vedclass · Accepted Answer

$171$

Vedclass · Answer

(A) Using the Binomial Theorem,the expansions are:$(1+2x)^{6} = \sum_{k=0}^{6} \binom{6}{k} (2x)^k = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6$$(1-x)^{7} = \sum_{r=0}^{7} \binom{7}{r} (-x)^r = 1 - 7x + 21x^2 - 35x^3 + 35x^4 - 21x^5 + 7x^6 - x^7$To find the coefficient of $x^5$ in the product,we multiply terms from the two expansions such that the sum of the powers of $x$ is $5$:$1 	imes (-21x^5) = -21x^5$$(12x) 	imes (35x^4) = 420x^5$$(60x^2) 	imes (-35x^3) = -2100x^5$$(160x^3) 	imes (21x^2) = 3360x^5$$(240x^4) 	imes (-7x) = -1680x^5$$(192x^5) 	imes (1) = 192x^5$Summing these coefficients: $-21 + 420 - 2100 + 3360 - 1680 + 192 = 171$Thus,the coefficient of $x^5$ is $171$.

Find the coefficient of $x^{5}$ in the product $(1+2x)^{6}(1-x)^{7}$ using the binomial theorem.

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