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{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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none of the above

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Solution

The correct option is A none of the above
$S_{n} = 1 + (1 + a) + (1 + a + a^{2}) + . . . . . . (1 - a) S_{n} = (1 - a) + (1 - a^{2}) + (1 - a^{3}) + (1 - a^{4}) + . . . . . (1 - a) S_{n} = (1 + 1 + 1 + . . . . . n) - (a + a^{2} + a^{3} + . . . . . .) (1 - a) S_{n} = n - \frac{a (a^{n} - 1)}{(a - 1)} S_{n} = \frac{n}{(1 - a)} + \frac{(a^{n} - 1) a}{{(1 - a)}^{2}}$

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Similar questions

a_{1}

a_{2}

a_{3}

,.....,

a_{n}

are in an

A . P .

\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}}}

Q. If

a_{1}, a_{2}, a_{3}, . . ., a_{n}

are in A.P of non-zero terms, Prove that

\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}}

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

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

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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 none of the aboveSn=1+(1+a)+(1+a+a2)+......(1−a)Sn=(1−a)+(1−a2)+(1−a3)+(1−a4)+.....(1−a)Sn=(1+1+1+.....n)−(a+a2+a3+......)(1−a)Sn=n−a(an−1)(a−1)Sn=n(1−a)+(an−1)a(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