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The infinite sum $1 + \frac{4}{7} + \frac{9}{7^2} + \frac{16}{7^3} + \frac{25}{7^4} + \dots$  equals

  1. $\frac{27}{14}$
  2. $\frac{21}{13}$
  3. $\frac{49}{27}$
  4. $\frac{256}{147}$
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S  = 1 + $\frac{4}{7}$ + $\frac{9}{7^{2}}$ + $\frac{16}{7^{3}}$ + $\frac{25}{7^{4}}$ +.................... (A)

$\frac{S}{7}$=       $\frac{1}{7}$ + $\frac{4}{7^{2}}$ + $\frac{9}{7^{3}}$ +    $\frac{16}{7^{4}}$ +.................... (B)


Substract equation (B) from (A),

$\frac{6S}{7}$ = 1 + $\frac{3}{7}$ + $\frac{5}{7^{2}}$ + $\frac{7}{7^{3}}$ + $\frac{9}{7^{4}}$  + ...................... (C)

$\frac{6S}{7^{2}}$=      $\frac{1}{7}$ + $\frac{3}{7^{2}}$ + $\frac{5}{7^{3}}$ + $\frac{7}{7^{4}}$  + ...................... (D)


Substract equation (D) from (C),

$\frac{36S}{49}$  = 1 + $\frac{2}{7}$ + $\frac{2}{7^{2}}$ + $\frac{2}{7^{3}}$ + $\frac{2}{7^{4}}$  + .................

$\frac{36S}{49}$  = 1 + $\frac{2 }{7}$ * $\frac{7 }{6}$

$\frac{36S}{49}$  = $\frac{4}{3}$

  S   = $\frac{4 * 49}{3* 36}$

  S   = $\frac{49}{27}$

Ans- C.

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