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प्रश्न
\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]is real to
पर्याय
\[\frac{\left( 2n - 1 \right) \times 3^n}{4}\]
\[\frac{\left( 2n - 1 \right) \times 3^n + 1}{4}\]
\[\left( 2n - 1 \right) 3^n + 1\]
\[\frac{\left( 2n - 1 \right) \times 3^n - 1}{4}\]
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उत्तर
\[\frac{\left( 2n - 1 \right) \times 3^n + 1}{4}\]
\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]
\[ = \lim_{x \to 3} \frac{x^1 + x^2 + x^3 + . . . . . . + x^n - \left( 3^1 + 3^2 + 3^3 . . . . . 3^n \right)}{x - 3}\]
\[ = \lim_{x \to 3} \frac{x - 3}{x - 3} + \frac{x^2 - 3^2}{x - 3} + \frac{x^3 - 3^3}{x - 3} + . . . . \frac{x^n - 3^n}{x - 3}\]
\[ = 1 + 2 \times 3 + 3 \times 3^2 + . . . . . + n 3^{n - 1} \left[ \because \frac{x^n - a^n}{x - a} = n a^{n - 1} \right]\]
This series is an AGP of the form given below:
S = 1 + 2r + 3r2........nrn–1
\[S = \frac{1 - r^n}{\left( 1 - r \right)^2} - \frac{n r^n}{1 - r}\]
\[r = 3, a = 1 d = 1\]
\[ = \frac{1 - 3^n + 2n 3^n}{4}\]
\[ = \frac{3^n \left( 2n - 1 \right) + 1}{4}\]
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