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:

& A.9 \\\ & B.27 \\\ & C.108 \\\ & D.54 \\\ \end{aligned}$$

Answer 1

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:

& {{\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{aligned}$$ Taking ${{\left( \dfrac{3}{4} \right)}^{3}}$ common from above we get: $$\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{aligned} & \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{aligned}$$ Dividing by 4 we get: $$\Rightarrow \dfrac{27}{16}\times 40\times 120$$ Again solving: $$\begin{aligned} & \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{aligned}$$ We were given that, the value of sum of series $${{\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.