Question

How do you evaluate the sum represented by $\sum{{}}$ with $n^3$ ? Any examples.

Answer 1

Here in this question we have been asked to evaluate the sum represented by $\sum{{}}$ with $n^3$. We need to show some examples. From the basics of algebra we know that we have a formula for $\sum{n}=\dfrac{n\left( n+1 \right)}{2}$ , $\sum{{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}$ and similarly we have a formula for $\sum{{{n}^{3}}}$ .

Complete step by step solution:
Now considering from the question we have been asked to evaluate the sum represented by $\sum{{}}$ with $n^3$. We need to show some examples.
From the basics of algebra we know that we have a formula for $\sum{n}=\dfrac{n\left( n+1 \right)}{2}$ , $\sum{{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}$ and similarly we have a formula for $\sum{{{n}^{3}}}$ which is given as $\sum{{{n}^{3}}}={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$ .
For example if we want to evaluate the sum of cubes of $1,2,3,4,5$ is given as $\begin{aligned} & \sum\limits_{n=1}^{5}{{{n}^{3}}}={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}} \\\ & \Rightarrow 1+8+27+64+125=225 \\\ \end{aligned}$ .
If we apply the formula then we will have $\Rightarrow {{\left( \dfrac{5\times 6}{2} \right)}^{2}}={{5}^{2}}\times {{3}^{2}}\Rightarrow 25\times 9=225$ .
Similarly we can evaluate the sum of cubes of $1,2,3,4,5,6,7,8,9,10$ then we will have $\Rightarrow {{\left( \dfrac{10\times 11}{2} \right)}^{2}}={{5}^{2}}\times {{11}^{2}}\Rightarrow 25\times 121=3025$ . We can evaluate this by doing the sum of cubes step by step but it is not an efficient way. So it is not performed.

Therefore we can conclude that the sum of cube of first $n$ numbers is given by $\sum{{{n}^{3}}}={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$.

Note: While answering questions of this type we should be sure with the calculations we perform and concepts we apply. If we are strong with our basic concept we can answer this question easily and in a short span of time. Similarly we can evaluate the sum of first $n$ numbers or the sum of first $n$ squares.