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Question: Find the value of \(1+\dfrac{3}{2}+\dfrac{5}{{{2}^{2}}}+\dfrac{7}{{{2}^{3}}}+\ldots \ldots \infty \)...

Find the value of 1+32+522+723+……∞1+\dfrac{3}{2}+\dfrac{5}{{{2}^{2}}}+\dfrac{7}{{{2}^{3}}}+\ldots \ldots \infty :
A. 3
B. 6
C. 9
D. 12

Explanation

Solution

For solving this question, first we will analyse the given series and write the expression of rth{{r}^{th}} term of the given series and then write in the form of Tr=f(r)−f(r+1){{T}_{r}}=f\left( r \right)-f\left( r+1 \right). After that we will apply the summation and find the sum of the first nn terms and then we will find the sum of infinite terms of the given series.

Complete step by step answer:
According to the question it is asked to us to find the sum of 1+32+522+723+……∞1+\dfrac{3}{2}+\dfrac{5}{{{2}^{2}}}+\dfrac{7}{{{2}^{3}}}+\ldots \ldots \infty . As we know that there are many methods to solve the sum of any series till infinite. But here we will use the shortest method for solving this. After evaluating the series, we can say that the terms are continuously decreasing after the second term of this series. It means at the last or at infinity it will be equal to near zero or we can say that at infinite it will tend to zero. So, we can use the sum of terms directly here. So, we can write as:
Let S=1+32+522+723+……∞………(i)S=1+\dfrac{3}{2}+\dfrac{5}{{{2}^{2}}}+\dfrac{7}{{{2}^{3}}}+\ldots \ldots \infty \ldots \ldots \ldots \left( i \right)
And if we divide this by 2, then we will get,
S2=12+322+523+724+……∞………(ii)\dfrac{S}{2}=\dfrac{1}{2}+\dfrac{3}{{{2}^{2}}}+\dfrac{5}{{{2}^{3}}}+\dfrac{7}{{{2}^{4}}}+\ldots \ldots \infty \ldots \ldots \ldots \left( ii \right)
Now if we subtract equation (ii) from (i), then we will get,
S−S2=1+22+222+223+224……∞ S2=1+(1+12+122+123……) \begin{aligned} & S-\dfrac{S}{2}=1+\dfrac{2}{2}+\dfrac{2}{{{2}^{2}}}+\dfrac{2}{{{2}^{3}}}+\dfrac{2}{{{2}^{4}}}\ldots \ldots \infty \\\ & \dfrac{S}{2}=1+\left( 1+\dfrac{1}{2}+\dfrac{1}{{{2}^{2}}}+\dfrac{1}{{{2}^{3}}}\ldots \ldots \right) \\\ \end{aligned}
Now sum of infinite GP can be written as follows,
=1+(11−12) =1+2=3 \begin{aligned} & =1+\left( \dfrac{1}{1-\dfrac{1}{2}} \right) \\\ & =1+2=3 \\\ \end{aligned}
So,
S=2×3=6S=2\times 3=6

So, the correct answer is “Option B”.

Note: Here the student should try to understand what is asked in the question. After that we should try to analyse the given series and somehow write the expression of rth{{r}^{th}} term. We should write it correctly without any mistake and then write in the form of Tr=f(r)−f(r+1){{T}_{r}}=f\left( r \right)-f\left( r+1 \right). So that we will be able to calculate the value of summation easily without any mistakes.