Question

See addition of the first n elements of the series 6 + 66 + 666 +​

Answer 1

Let S be the sum of series 6+66+666+...to n terms.

Then ,

$\sf \: S = 6 + 66 + 666 + ...to \: n \: terms$

$\sf S = 6 \big\{1 + 11 + 111 + ...to \: n \: terms \big\}$

$\sf \: S = \dfrac{6}{9} \times 9 \big\{1 + 11 + 111 + ...to \: n \: terms \big\}$

$\sf \: S = \dfrac{6}{9} \big\{9 + 99 + 999 + ...to \: n \: terms \big\}$

$\sf \: S = \dfrac{6}{9} \bigg\{ \big(10 - 1 \big) + \big( {10}^{2} - 1 \big) + \big( {10}^{3} - 1 \big) + ... \big( {10}^{n} - 1 \big) \bigg\}$

$\sf \: S = \dfrac{6}{9} \bigg \{ \big(10 + {10}^{2} + {10}^{3} + ... + {10}^{n} \big) - \big(1 + 1 + 1 + ...n \: terms \big) \bigg \}$

$\sf \: S \: = \dfrac{6}{9} \bigg \{10 \times \dfrac{ \big( {10}^{n - 1} \big)}{10 - 1} - n \bigg \}$

$\sf \: S = \dfrac{6}{9} \bigg \{ \dfrac{10 \times \big( {10}^{n} - 1 \big) - 9n }{9} \bigg \}$

$\sf \: S \: = \dfrac{6}{81} \big\{ {10}^{n + 1} - 10 - 9n \big\}$

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