The sum of n terms of the series $1 + (1 + a) + (1 + a + a^{2}) + (1 + a + a^{2} + a^{3}) + . . . . .$ , is

\frac{n}{1 - a} - \frac{a (1 - a^{n})}{(1 - a)^{2}}

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\frac{n}{1 - a} + \frac{a (1 - a^{n})}{(1 - a)^{2}}

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\frac{n}{1 - a} + \frac{a (1 + a^{n})}{(1 - a)^{2}}

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\frac{n}{1 - a} - \frac{a (1 + a^{n})}{(1 - a)^{2}}

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Solution

The correct option is A $\frac{n}{1 - a} - \frac{a (1 - a^{n})}{(1 - a)^{2}}$
$T_{n} = 1 + a + a^{2} + . . . . + a^{n - 1} = \frac{1 - a^{n}}{1 - a}$
$S n = \sum T_{n} = \frac{1}{(1 - a)} [\sum 1 - \sum a^{n}]$
$= \frac{1}{(1 - a)} [n - (a - a^{2} + a^{3} + . . . . a^{n})]$
$= \frac{1}{(1 - a)} [n - \frac{a (1 - a^{n})}{(1 - a)}] = \frac{n}{1 - a} - \frac{a (1 - a^{n})}{(1 - a)^{2}}$

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\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{(n - 1)}{\sqrt{a_{n}} + \sqrt{a_{1}}}

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\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + . . . \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}

$a_{1}, a_{2}, \dots \dots . a_{n}$ are A.P. Where $a_{1}$ > 0 for all i, then $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}}$

If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =$

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The sum of n terms of the series 1+(1+a)+(1+a+a2)+(1+a+a2+a3)+....., is

The correct option is A n1−a−a(1−an)(1−a)2Tn=1+a+a2+....+an−1=1−an1−aSn=∑Tn=1(1−a)[∑1−∑an]=1(1−a)[n−(a−a2+a3+....an)]=1(1−a)[n−a(1−an)(1−a)]=n1−a−a(1−an)(1−a)2

The sum of n terms of the series $1 + (1 + a) + (1 + a + a^{2}) + (1 + a + a^{2} + a^{3}) + . . . . .$ , is