It is given that \({{x}^{5}}{{\left( x+3 \right)}^{3}}={{a}_{8}}{{\left( x+1 \right)}^{8}}+{{a}

Question

It is given that ${{x}^{5}}{{\left( x+3 \right)}^{3}}={{a}_{8}}{{\left( x+1 \right)}^{8}}+{{a}_{7}}{{\left( x+1 \right)}^{7}}+\cdots +{{a}_{1}}\left( x+1 \right)+{{a}_{0}}$. By differentiating both side with respect to $x$, show that ${{x}^{4}}{{\left( x+3 \right)}^{2}}\left( 8x+15 \right)=8{{a}_{8}}{{\left( x+1 \right)}^{7}}+7{{a}_{7}}{{\left( x+1 \right)}^{6}}+\cdots +{{a}_{1}}$. Hence, or otherwise, find the value of $7{{a}_{7}}+5{{a}_{5}}+3{{a}_{3}}+{{a}_{1}}$.

Accepted Answer

(i) $x^4(x+3)^2(8x+15)=8a_8(x+1)^7+7a_7(x+1)^6+\cdots+a_1$ (ii) $7a_7+5a_5+3a_3+a_1=-8$

Binomial Theorem - Sums of Coefficients in Polynomial Expansions

Binomial Theorem - Sums of Coefficients in Polynomial Expansions

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