Question

The sum of all numbers between 100 and 10,000 which are of the form ${{n}^{3}}\left( n\in N \right)$ is equal to
(a) 55216
(b) 53261
(c) 51261
(d) None of these

Answer 1

Firstly, we have to find the least and highest values of n that come nearest to 100 and 10000 respectively, when cubed. We will get the same as 5 and 21. Then, we have to add the cubes of all the numbers from 5 to 21. Then, we have to add and subtract the sum of cubes of numbers from 1 to 4. Finally, we have to apply the formula for the sum of cubes of n natural numbers which is given by $\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$ and simplify.

Complete step by step answer:
We have to find the sum of all numbers between 100 and 10,000 which are of the form ${{n}^{3}}$ . We know that the least value of n that comes nearest to 100, when cubed is 5.
$\Rightarrow {{5}^{3}}=125$
We know that the highest value of n that equals to or comes below 10,000 when cubed is 21.
$\Rightarrow {{21}^{3}}=9261$
Now, we can write the sum of all numbers between 100 and 10,000 which are of the form ${{n}^{3}}$ as
$\Rightarrow {{5}^{3}}+{{6}^{3}}+{{7}^{3}}+...+{{21}^{3}}$
We have to rewrite the above expression by adding and subtracting ${{1}^{3}}+{{2}^{3}}+{{3}^{2}}+{{4}^{3}}$ .

& \Rightarrow \left( {{1}^{3}}+{{2}^{3}}+{{3}^{2}}+{{4}^{3}} \right)+{{5}^{3}}+{{6}^{3}}+...+{{21}^{3}}-\left( {{1}^{3}}+{{2}^{3}}+{{3}^{2}}+{{4}^{3}} \right) \\\ & \Rightarrow \left( {{1}^{3}}+{{2}^{3}}+{{3}^{2}}+{{4}^{3}}+{{5}^{3}}+{{6}^{3}}+...+{{21}^{3}} \right)-\left( {{1}^{3}}+{{2}^{3}}+{{3}^{2}}+{{4}^{3}} \right) \\\ \end{aligned}$$ We can write the above sum as $\Rightarrow \sum\limits_{n=1}^{21}{{{n}^{3}}}-\sum\limits_{k=1}^{4}{{{k}^{3}}}...\left( i \right)$ We know that sum of cubes of n natural numbers are given by the formula $\Rightarrow {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$ Let us use the above formula in (i). $\begin{aligned} & \Rightarrow \left[ \dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}\text{ for }n=21 \right]-\left[ \dfrac{{{k}^{2}}{{\left( k+1 \right)}^{2}}}{4}\text{ for }n=4 \right] \\\ & =\dfrac{{{21}^{2}}{{\left( 21+1 \right)}^{2}}}{4}-\dfrac{{{4}^{2}}{{\left( 4+1 \right)}^{2}}}{4} \\\ & =\dfrac{{{21}^{2}}\times {{22}^{2}}}{4}-\dfrac{{{4}^{2}}\times {{5}^{2}}}{4} \\\ \end{aligned}$ We have to cancel the common factors from the numerator and denominator of each term. $\begin{aligned} & \Rightarrow \dfrac{21\times 21\times {{\require{cancel}\cancel{22}}^{11}}\times {{\require{cancel}\cancel{22}}^{11}}}{{{\require{cancel}\cancel{4}}^{\require{cancel}\cancel{2}}}}-\dfrac{{{4}^{\require{cancel}\cancel{2}}}\times {{5}^{2}}}{\require{cancel}\cancel{4}} \\\ & =21\times 21\times 11\times 11-4\times 5\times 5 \\\ & =53361-100 \\\ & =53261 \\\ \end{aligned}$ **So, the correct answer is “Option b”.** **Note:** Students must be thorough with the formulas of sum of $n,{{n}^{2}},{{n}^{3}},{{n}^{4}}$ . They have a chance of making mistake by writing the formula for the sum of cubes of n natural numbers as $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ which is the formula for the the sum of squares of n natural numbers.