Question
Find the coefficients of $a^4$ in the product $(1+2\text{a})^{4}(2-\text{a})^{5}$ using binomial theorem.

Answer

We have,
$(1+2\text{a})^{4}(2-\text{a})^{5}$
$=\Big[{^\text{4}}\text{C}_{\text{0}}\big(2\text{a}\big)^{0}+{^\text{4}}\text{C}_{\text{1}}\big(2\text{a}\big)^1+{^\text{4}}\text{C}_{\text{2}}\big(2\text{a}\big)^{2}+{^\text{4}}\text{C}_{\text{3}}\big(2\text{a}\big)^{3}+{^\text{4}}\text{C}_{\text{4}}\big(2\text{a}\big)^{4}\Big]\\ \times\Big[{^\text{5}}\text{C}_{\text{0}}(2)^{5}(-\text{a})^{0}+{^\text{5}}\text{C}_{\text{1}}(2)^{4}(-\text{a})^{1}+{^\text{5}}\text{C}_{\text{2}}(2)^{3}(-\text{a})^{2}+{^\text{5}}\text{C}_{\text{2}}(2)^{2}(-\text{a})^{3}+{^\text{5}}\text{C}_{\text{1}}(2)^{1}(-\text{a})^{5}$
$=\Big[1+8\text{a}+24\text{a}^{2}+32\text{a}^{3}+16\text{a}^{4}\Big]\times\Big[32-80\text{a}+80\text{a}^{2}-40\text{a}^{3}+10\text{a}^{4}-\text{a}^{5}\Big]$
Coefficient of $\text{a}^{4}=10-320+1920-2560+512=-438$

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