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Question Number 120016 by aurpeyz last updated on 28/Oct/20

Answered by mr W last updated on 28/Oct/20

$$S=\frac{{(1+x)}^{16}-{(1+x)}^6}{x}$$$$C_7^{16}=C_9^{16}$$$$\Rightarrow answer\left(a\right)$$

Answered by mathmax by abdo last updated on 29/Oct/20

$${(1+\mathrm{x})}^6+{(1+\mathrm{x})}^7+....{(1+\mathrm{x})}^{15}={(1+\mathrm{x})}^6\{1+(1+\mathrm{x})+{(1+\mathrm{x})}^2+...+{(1+\mathrm{x})}^9)$$$$={(1+\mathrm{x})}^6\times \frac{1-{(1+\mathrm{x})}^{10}}{-\mathrm{x}}={(1+\mathrm{x})}^6\times \frac{{(1+\mathrm{x})}^{10}-1}{\mathrm{x}}$$$$=\frac{{(1+\mathrm{x})}^{16}-{(1+\mathrm{x})}^6}{\mathrm{x}}=\frac{1}{\mathrm{x}}\{\sum _{\mathrm{k}=0}^{16}{\mathrm{C}}_{16}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}-\sum _{\mathrm{k}=0}^6{\mathrm{C}}_6^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}\}$$$$=\frac{1}{\mathrm{x}}(\sum _{\mathrm{k}=1}^{16}{\mathrm{C}}_{16}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}-\sum _{\mathrm{k}=1}^6{\mathrm{C}}_6^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}})=\sum _{\mathrm{k}=1}^{16}{\mathrm{C}}_{16}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}-1}-\sum _{\mathrm{k}=1}^6{\mathrm{C}}_6^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}-1}\Rightarrow \mathrm{the}\mathrm{coefficient}\mathrm{is}$$$$\lambda ={\mathrm{C}}_{16}^7=\frac{16!}{7!9!}$$

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