The sum of n terms of the series +1+4/5+{7}{5²}+{10}{5³}+ is

Correct Answer is : 54+1516(1−15n−1)−(3n−2)4⋅5n−15/4+15/16≤ft(1-{1}{5ⁿ-1}})-{(3n-2)}{4·5ⁿ-1}}45+1615(1−5n−11)−4⋅5n−1(3n−2)

5/4+15/16≤ft(1-{1}{5ⁿ-1}})-{(3n-2)}{4·5ⁿ-1}}

5/4+1/16≤ft(1-{1}{5ⁿ-1}})-{3n}{4·5ⁿ-1}}

≤ft(1-{1}{5ⁿ-1}})-{(3n+2)}{4·5ⁿ-1}}

Correct Answer is : 54+1516(1−15n−1)−(3n−2)4⋅5n−15/4+15/16≤ft(1-{1}{5ⁿ-1}})-{(3n-2)}{4·5ⁿ-1}}45+1615(1−5n−11)−4⋅5n−1(3n−2)

Test Index

Manipal Entrance Test 2013 Math Paper

Question : 3 of 80

Marks: +1, -0

The sum of

n

+1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+\dots

$\frac{5}{4}+\frac{15}{16}\left(1-\frac{1}{5^{n-1}}\right)-\frac{(3n-2)}{4\cdot5^{n-1}}$
$\frac{5}{4}+\frac{1}{16}\left(1-\frac{1}{5^{n-1}}\right)-\frac{3n}{4\cdot5^{n-1}}$
$\left(1-\frac{1}{5^{n-1}}\right)-\frac{(3n+2)}{4\cdot5^{n-1}}$
None of the above

Solution:

Clearly, the given series is an arithmetico-geometric series whose corresponding AP and GP are respectively.

1 ,4 ,7 ,1 0 , and

1, \frac{1}{5}, \frac{1}{5^{2}}, \frac{1}{5^{3}}, \dots

The

n^{\text{th}}

term of A P

=[1+(n-1)\times 3]=3n-2

The

n^{\text{th}}

term of G P

=\left[1\times\left(\frac{1}{5}\right)^{n-1}\right]=\left(\frac{1}{5}\right)^{n-1}

So the

n^{\text{th}}

term of the given series is

(3n-2)\times\frac{1}{5^{n-1}}=\frac{3n-2}{5^{n-1}}

Let,

S_{n}=1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+\dots+\frac{3n-5}{5^{n-2}}+\frac{3n-2}{5^{n-1}}

......(i)

\frac{1}{5}S_{n}=\frac{1}{5}+\frac{4}{5^{2}}+\frac{7}{5^{3}}+\dots+\frac{(3n-5)}{5^{n-1}}+\frac{3n-2}{5^{n}}

....(ii)

Subtracting (ii) from (i), we get

S_{n}-\frac{1}{5}S_{n}=1+\left\{\frac{3}{5}+\frac{3}{5^{2}}+\frac{3}{5^{3}}+\dots+\frac{3}{5^{n-1}}\right\}-\frac{(3n-2)}{5^{n}}

⇒

\frac{4}{5}S_{n}=1+\frac{3}{5}\frac{\left\{1-\left(\frac{1}{5}\right)^{n-1}\right\}}{\left(1-\frac{1}{5}\right)}-\frac{(3n-2)}{5^{n}}

⇒

\frac{4}{5}S_{n}=1+\frac{3}{5}\frac{\left\{1-\frac{1}{5^{n-1}}\right\}}{\left(\frac{4}{5}\right)}-\frac{(3n-2)}{5^{n}}

⇒

\frac{4}{5}S_{n}=1+\frac{3}{4}\left(1-\frac{1}{5^{n-1}}\right)-\frac{(3n-2)}{5^{n}}

⇒

S_{n}=\frac{5}{4}+\frac{15}{16}\left(1-\frac{1}{5^{n-1}}\right)-\frac{(3n-2)}{4\cdot5^{n-1}}

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