The sum of 1 - 2-5 - 3-5-2 - 4-5-3 - - upto n-terms is -MP PET 1982-25-16 - 4n - 5-16 - 5-n-1-3-4 - 2n - 5-16 - 5-n-1-3-7 - 3n - 5-16 - 5-n-1-1-2 - 5n - 1-3 - 5-n-2-

The correct option is A 25/16 4n + 5/16 × 5^n 1.Given series, let S_n = 1 + 2/5 + 3/5^2 + 4/5^3 + ....... + n/5^n 1 1/5 S_n = 1/5 + 2/5^2 + 3/5^3 + ....... + ...

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Byju's Answer

Standard X

Mathematics

Nth term of A.G.P

The sum of 1 ...

Question

The sum of $1 + \frac{2}{5} + \frac{3}{5^{2}} + \frac{4}{5^{3}} + . . . . . . .$ upto n terms is
[MP PET 1982]

\frac{25}{16} - \frac{4 n + 5}{16 \times 5^{n - 1}} .

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\frac{3}{4} - \frac{2 n + 5}{16 \times 5^{n + 1}} .

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\frac{3}{7} - \frac{3 n + 5}{16 \times 5^{n - 1}} .

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\frac{1}{2} - \frac{5 n + 1}{3 \times 5^{n + 2}} .

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Solution

The correct option is A $\frac{25}{16} - \frac{4 n + 5}{16 \times 5^{n - 1}} .$
Given series, let $S_{n} = 1 + \frac{2}{5} + \frac{3}{5^{2}} + \frac{4}{5^{3}} + . . . . . . . + \frac{n}{5^{n - 1}}$
$\frac{1}{5} S_{n} = \frac{1}{5} + \frac{2}{5^{2}} + \frac{3}{5^{3}} + . . . . . . . + \frac{n}{5^{n}}$
Subtracting, $(1 - \frac{1}{5}) S_{n} = 1 + \frac{1}{5} + \frac{2}{5^{2}} + \frac{3}{5^{3}} + . . . . . + \frac{n}{5^{n}}$
$\Rightarrow \frac{4}{5} S_{n} = \frac{1 - \frac{1}{5^{n}}}{\frac{4}{5}} - \frac{n}{5^{n}} \Rightarrow S_{n} = \frac{25}{16} - \frac{4 n + 5}{16 \times 5^{n - 1}} .$

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The sum of 1+25+352+453+....... upto n terms is [MP PET 1982]

The correct option is A 2516−4n+516×5n−1.Given series, let Sn=1+25+352+453+.......+n5n−1 15Sn=15+252+353+.......+n5n Subtracting, (1−15)Sn=1+15+252+353+.....+n5n ⇒45Sn=1−15n45−n5n⇒Sn=2516−4n+516×5n−1.

The sum of $1 + \frac{2}{5} + \frac{3}{5^{2}} + \frac{4}{5^{3}} + . . . . . . .$ upto n terms is
[MP PET 1982]