Find 2 + 22 + 222 + 2222 + …… upto n terms. - Vidyadip

Question

Find 2 + 22 + 222 + 2222 + …… upto n terms.

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Answer

$
\begin{aligned}
& \mathrm{S}_{\mathrm{n}}=2+22+222+\ldots . \text { upto } \mathrm{n} \text { terms } \\
& =2(1+11+111+\ldots . . \text { upto } \mathrm{n} \text { terms }) \\
& =\frac{2}{9}(9+99+999+\ldots \text { upto } \mathrm{n} \text { terms }) \\
& =\frac{2}{9}[(10-1)+(100-1)+(1000-1)+\ldots . . \text { upto } \mathrm{n} \text { terms }] \\
& =\frac{2}{9}[(10+100+1000+\ldots \text { upto } \mathrm{n} \text { terms })-(1+1+1+\ldots . \mathrm{n} \text { times })]
\end{aligned}
$
Since, $10,100,1000, \ldots . . n$ terms are in G.P.
with $a=10, r=\frac{100}{10}=10$
$
\therefore \quad \mathrm{S}_{\mathrm{n}}=\frac{2}{9}\left[10\left(\frac{10^{\mathrm{n}}-1}{10-1}\right)-\mathrm{n}\right]=\frac{2}{9}\left[\frac{10}{9}\left(10^{\mathrm{n}}-1\right)-\mathrm{n}\right]
$
$
\therefore \quad \mathrm{S}_{\mathrm{n}}=\frac{2}{81}\left[10\left(10^{\mathrm{n}}-1\right)-9 \mathrm{n}\right]
$

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