MCQ
Value of $\sum\limits_{n = 0}^\infty {\frac{{{{(n + 1)}^2}}}{{{7^n}}}}$ is -
- ✓$\frac{49}{27}$
- B$\frac{27}{49}$
- C$\frac{21}{13}$
- D$\frac{27}{14}$
$\frac{\frac{s}{7}=\frac{1^{2}}{7}+\frac{2^{2}}{7^{2}}+\frac{3^{2}}{7^{3}}+\ldots \infty}{\frac{6 s}{7}=1+\frac{3}{7}+\frac{5}{7^{2}}+\frac{9}{7^{3}}+\ldots \ldots \infty}$
$\frac{6 s}{7^{2}}=\frac{1}{7}+\frac{3}{7^{2}}+\frac{5}{7^{3}}+\frac{9}{7^{4}}+\ldots \ldots \infty$
$\therefore \frac{36 s}{7^{2}}=1+\frac{2}{7}+\frac{2}{7^{2}}+\frac{2}{7^{3}}+\ldots . . \infty$
$\Rightarrow \frac{36 s}{49}=\frac{4}{3} \Rightarrow s=\frac{49}{27}$
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