Question

If the sum of the first 15 terms of the series \[{{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( 1\dfrac{1}{2} \right)}^{3}}+{{\left( 2\dfrac{1}{4} \right)}^{3}}+{{3}^{3}}+{{\left( 3\dfrac{3}{4} \right)}^{3}}+......\] is equal to 225 k. then k is equal to:
\[\begin{align}
  & A.9 \\
 & B.27 \\
 & C.108 \\
 & D.54 \\
\end{align}\]

Answer 1

Hint: To solve this question, we will first convert mixed fraction to fraction which are given in series \[{{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( 1\dfrac{1}{2} \right)}^{3}}+{{\left( 2\dfrac{1}{4} \right)}^{3}}+{{3}^{3}}+{{\left( 3\dfrac{3}{4} \right)}^{3}}+......\]
Then we will try to find common differences between terms by subtracting consecutive terms. If the common difference is the same, then the term can be written as a product of the common difference. Finally, we will use formula of sum of cube of n terms which is ${{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$

Complete step-by-step answer:
We are given the series as
\[{{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( 1\dfrac{1}{2} \right)}^{3}}+{{\left( 2\dfrac{1}{4} \right)}^{3}}+{{3}^{3}}+{{\left( 3\dfrac{3}{4} \right)}^{3}}+......\]
Converting the mixed fraction to fraction, we get series as:
\[{{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( \dfrac{3}{2} \right)}^{3}}+{{\left( \dfrac{9}{4} \right)}^{3}}+{{3}^{3}}+{{\left( \dfrac{15}{4} \right)}^{3}}+......\]
Subtracting the second term $\dfrac{3}{2}$ to $\dfrac{3}{4}$ first term we get:
\[\Rightarrow \left( \dfrac{3}{2} \right)-\left( \dfrac{3}{4} \right)=\dfrac{3}{2}-\dfrac{3}{4}=\dfrac{6-3}{4}=\dfrac{3}{4}\]
So, the difference of first and second term is $\dfrac{3}{4}$
Similarly, the difference of third and second term is:
\[\Rightarrow \left( \dfrac{9}{4} \right)-\left( \dfrac{3}{2} \right)=\dfrac{9}{4}-\dfrac{3}{2}=\dfrac{9-6}{4}=\dfrac{3}{4}\]
Again difference is $\dfrac{3}{4}$
Hence, we see for all the consecutive terms of the given series \[{{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( \dfrac{3}{2} \right)}^{3}}+{{\left( \dfrac{9}{4} \right)}^{3}}+{{3}^{3}}+{{\left( \dfrac{15}{4} \right)}^{3}}+......\] the difference is $\dfrac{3}{4}$
Then, writing it as multiple of $\dfrac{3}{4}$ we get the series as:
\[\begin{align}
  & {{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( \dfrac{3}{2} \right)}^{3}}+{{\left( \dfrac{9}{4} \right)}^{3}}+{{3}^{3}}+{{\left( \dfrac{15}{4} \right)}^{3}}+...... \\
 & \Rightarrow {{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( 2\times \dfrac{3}{4} \right)}^{3}}+{{\left( 3\times \dfrac{3}{4} \right)}^{3}}+{{\left( 4\times \dfrac{3}{4} \right)}^{3}}+{{\left( 5\times \dfrac{3}{4} \right)}^{3}}+....... \\
\end{align}\]
Taking ${{\left( \dfrac{3}{4} \right)}^{3}}$ common from above we get:
\[\Rightarrow {{\left( \dfrac{3}{4} \right)}^{3}}\left\{ 1+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}+... \right\}\]
Also, as we are given that, we have 15 terms in the above series. So, last term would be ${{\left( \dfrac{3}{4} \right)}^{3}}\times 15$
Hence, we have
\[\Rightarrow {{\left( \dfrac{3}{4} \right)}^{3}}\left\{ 1+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}+{{...15}^{3}} \right\}\]
Opening cube of 3 and 4 we get:
\[\Rightarrow \left( \dfrac{27}{64} \right)\left\{ 1+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}+{{...15}^{3}} \right\}\]
We know that, the sum of cube of n number is given as:
\[1+{{2}^{3}}+...{{n}^{3}}={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}\]
Using this in \[\left\{ 1+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}+{{...15}^{3}} \right\}\] using n is 15 we get:
\[\Rightarrow \left( \dfrac{27}{64} \right)={{\left( \dfrac{15\left( 15+1 \right)}{2} \right)}^{2}}\]
We have
\[\begin{align}
  & \Rightarrow \left( \dfrac{27}{64} \right)={{\left( \dfrac{15\times 16}{2} \right)}^{2}} \\
 & \Rightarrow \left( \dfrac{27}{64} \right)={{\left( 120 \right)}^{2}} \\
 & \Rightarrow \dfrac{27}{64}\times 120\times 120 \\
\end{align}\]
Dividing by 4 we get:
\[\Rightarrow \dfrac{27}{16}\times 40\times 120\]
Again solving:
\[\begin{align}
  & \Rightarrow 27\times 225 \\
 & \Rightarrow 27\times 225\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
\end{align}\]
We were given that, the value of sum of series \[{{\left( \dfrac{3}{4} \right)}^{3}}+{{\left( \dfrac{3}{2} \right)}^{3}}+{{\left( \dfrac{9}{4} \right)}^{3}}+{{3}^{3}}+{{\left( \dfrac{15}{4} \right)}^{3}}+......\] is 225 k
Comparing it with equation (i) we have:
\[\Rightarrow 225\times 27=225\text{ k}\]
Dividing 225 both sides,
\[\therefore \text{k = 225}\]
Therefore, the value of k is 27

So, the correct answer is “Option B”.

Note: The key point to note in this question is that, while you are opening the bracket of sum of series \[\Rightarrow {{\left( \dfrac{3}{4} \right)}^{3}}\left\{ 1+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}+{{...15}^{3}} \right\}\] then do not arrive at a general answer. Try to form a product of the type 225 k. Because the sum is 225 k then, we can easily compare the value of k obtained to get the result.
Anyway solving the sum value and then calculating k by dividing by 225 is also correct but it increases the calculation steps.