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Question: The sum of the infinite series \[1 + 2 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} +...

The sum of the infinite series 1+2+23+632+1033+1434+.......1 + 2 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + ....... is:
A) 6
B) 92\dfrac{9}{2}
C) 4
D) 5

Explanation

Solution

Here we will first assume the fractional part of the infinite series as a variable. Then we will find the value of that variable by converting it in terms of the sum of the G.P. series. We will use the basic formula of the sum of the G.P. series to get the value of that variable. Then by using the value of that variable we will get the value of the sum of the given infinite series.

Formula used:
We will use the formula of sum of an infinite G.P. series S=a1r{S_\infty } = \dfrac{a}{{1 - r}}, where aa is the first term and rr is the common ratio.

Complete Step by Step Solution:
Given series is 1+2+23+632+1033+1434+.......1 + 2 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + .......
Let S=1+2+23+632+1033+1434+.......S = 1 + 2 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + .......
Adding the first two terms, we get
S=3+23+632+1033+1434+.......\Rightarrow S = 3 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + .......
We can write the above equation as
S=3+L\Rightarrow S = 3 + L……………………….(1)\left( 1 \right)
Here, L=23+632+1033+1434+.......L = \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + ....... …………………….(2)\left( 2 \right)

Now we will take the expression of LL and find its value.
We will divide the expression of LL by 3. Therefore, we get
L3=232+633+1034+1435+.......\dfrac{L}{3} = \dfrac{2}{{{3^2}}} + \dfrac{6}{{{3^3}}} + \dfrac{{10}}{{{3^4}}} + \dfrac{{14}}{{{3^5}}} + .......……………………….(3)\left( 3 \right)
Now we will subtract the equation (3)\left( 3 \right) from equation (2)\left( 2 \right). Therefore, we get
LL3=(23+632+1033+1434+.......)(232+633+1034+1435+.......)\Rightarrow L - \dfrac{L}{3} = \left( {\dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + .......} \right) - \left( {\dfrac{2}{{{3^2}}} + \dfrac{6}{{{3^3}}} + \dfrac{{10}}{{{3^4}}} + \dfrac{{14}}{{{3^5}}} + .......} \right)
2L3=(23+632232+1033633+14341034+.......)\Rightarrow \dfrac{{2L}}{3} = \left( {\dfrac{2}{3} + \dfrac{6}{{{3^2}}} - \dfrac{2}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} - \dfrac{6}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} - \dfrac{{10}}{{{3^4}}} + .......} \right)
Simplifying the above equation, we get
2L3=(23+432+433+434+.......)\Rightarrow \dfrac{{2L}}{3} = \left( {\dfrac{2}{3} + \dfrac{4}{{{3^2}}} + \dfrac{4}{{{3^3}}} + \dfrac{4}{{{3^4}}} + .......} \right)
2L3=23+4(132+133+134+.......)\Rightarrow \dfrac{{2L}}{3} = \dfrac{2}{3} + 4\left( {\dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + \dfrac{1}{{{3^4}}} + .......} \right)
We can clearly see that the term (132+133+134+.......)\left( {\dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + \dfrac{1}{{{3^4}}} + .......} \right) is a sum of a G.P. series with first term a=132a = \dfrac{1}{{{3^2}}} and common ratio r=13r = \dfrac{1}{3}. so, we will use the formula of the sum of the G.P. series to get its value.
Using the formula, S=a1r{S_\infty } = \dfrac{a}{{1 - r}}, we get
2L3=23+4(132113)\Rightarrow \dfrac{{2L}}{3} = \dfrac{2}{3} + 4\left( {\dfrac{{\dfrac{1}{{{3^2}}}}}{{1 - \dfrac{1}{3}}}} \right)
Subtracting the terms in the denominator, we get
2L3=23+4(13223)\Rightarrow \dfrac{{2L}}{3} = \dfrac{2}{3} + 4\left( {\dfrac{{\dfrac{1}{{{3^2}}}}}{{\dfrac{2}{3}}}} \right)
2L3=23+4(16)\Rightarrow \dfrac{{2L}}{3} = \dfrac{2}{3} + 4\left( {\dfrac{1}{6}} \right)
Now we will solve this equation to get the value of LL. Therefore, we get
L=32(23+23)\Rightarrow L = \dfrac{3}{2}\left( {\dfrac{2}{3} + \dfrac{2}{3}} \right)
L=32(43)=2\Rightarrow L = \dfrac{3}{2}\left( {\dfrac{4}{3}} \right) = 2
Now we will put the value of LL in the equation (1)\left( 1 \right) to get the value of the sum. Therefore, we get
S=3+2=5\Rightarrow S = 3 + 2 = 5

Hence, the sum of the infinite series 1+2+23+632+1033+1434+.......1 + 2 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + ....... is 5.

Note:
Here we expressed the given series in GP or geometric progression. Geometric progression is defined as the sequence or the series in which the consecutive terms differ by a common ratio. Other than geometric progression, we also have different progression such as arithmetic progression and harmonic progression. Harmonic progression is defined as the sequence or the series in which the consecutive terms have a common difference. Harmonic progression is the reciprocal of arithmetic progression.