Gujarati
8. Sequences and Series
medium

अनुक्रम $3 + 33 + 333 + ....$ के $n$ पदों का योग होगा

A

$\frac{1}{{27}}({10^{n + 1}} + 9n - 28)$

B

$\frac{1}{{27}}({10^{n + 1}} - 9n - 10)$

C

$\frac{1}{{27}}({10^{n + 1}} + 10n - 9)$

D

इनमें से कोई नहीं

Solution

(b) श्रेणी  $3 + 33 + 333 +………+ n$ पदों तक

$ = \frac{1}{3}[9 + 99 + 999 + …….. + n{\rm{]}}$ पदों तक

$ = \frac{1}{3}[(10 – 1) + ({10^2} – 1) + ({10^3} – 1) + …. + n]$ पदों तक

$ = \frac{1}{3}[10 + {10^2} + …. + {10^n}]$$ – \frac{1}{3}[1 + 1 + 1 + …. + n]$ पदों तक

$ = \frac{1}{3}\,.\,\frac{{10\,({{10}^n} – 1)}}{{10 – 1}} – \frac{1}{3}.n\,$ $ = \frac{1}{3}\left[ {\frac{{{{10}^{n + 1}} – 10}}{9} – n} \right]$

$ = \frac{1}{3}\,\left[ {\frac{{{{10}^{n\, + \,1}} – 9n – 10}}{9}} \right]$ $ = \frac{1}{{27}}[{10^{n\, + \,1}} – 9n – 10]$

Standard 11
Mathematics

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