Question
Find the sum to n terms of the sequences $8, 88, 888, 8888$, ……
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Answer
Here $Sn = 8 + 88 + 888 + 8888 +$ ....... up to $n$ terms$\Rightarrow Sn = 8(1 +11 + 111 + 1111 +$ ...... up to $n$ terms)
$\Rightarrow S _ { n } = \frac { 8 } { 9 }$(9 + 99 + 999 + 9999 + ....... up to n terms)
$\Rightarrow S _ { n } = \frac { 8 } { 9 } \left[ ( 10 - 1 ) + \left( 10 ^ { 2 } - 1 \right) + \left( 10 ^ { 3 } - 1 \right) + \ldots . \text { up to } n \text { terms } \right]$
$\Rightarrow \mathrm { S } _ { n } = \frac { 8 } { 9 }[(10 + 10^2 + 10^3 +$ ..... up to $n$ terms) - (1 + 1 + 1 + ...... up to n terms)]
$\Rightarrow S _ { n } = \frac { 8 } { 9 } \left[ \frac { 10 \times \left( 10 ^ { n } - 1 \right) } { 10 - 1 } - n \right]$
$= \frac { 8 } { 9 } \left[ \frac { 10 } { 9 } \left( 10 ^ { n } - 1 \right) - n \right]$
$= \frac { 80 } { 81 } \left( 10 ^ { n } - 1 \right) - \frac { 8 } { 9 } n$
$\Rightarrow S _ { n } = \frac { 8 } { 9 }$(9 + 99 + 999 + 9999 + ....... up to n terms)
$\Rightarrow S _ { n } = \frac { 8 } { 9 } \left[ ( 10 - 1 ) + \left( 10 ^ { 2 } - 1 \right) + \left( 10 ^ { 3 } - 1 \right) + \ldots . \text { up to } n \text { terms } \right]$
$\Rightarrow \mathrm { S } _ { n } = \frac { 8 } { 9 }[(10 + 10^2 + 10^3 +$ ..... up to $n$ terms) - (1 + 1 + 1 + ...... up to n terms)]
$\Rightarrow S _ { n } = \frac { 8 } { 9 } \left[ \frac { 10 \times \left( 10 ^ { n } - 1 \right) } { 10 - 1 } - n \right]$
$= \frac { 8 } { 9 } \left[ \frac { 10 } { 9 } \left( 10 ^ { n } - 1 \right) - n \right]$
$= \frac { 80 } { 81 } \left( 10 ^ { n } - 1 \right) - \frac { 8 } { 9 } n$
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