Question

Find the sum of the cubes of the first five natural numbers. Verify your answer by using formula ${\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}$, where n is a natural number. Also find the sum of the cubes of the first ten natural numbers.

Answer 1

Hint: First we can individually find the cube of each natural number and then we can add all of them to find the sum of the first n natural numbers. Otherwise, we can use the formula to find the sum of cube of first n natural numbers but adding them individually is a long and tedious process.

Complete step-by-step answer:
In the given question,
First, we have to find the sum of the cubes of the first five natural numbers.
We know that,
${\left( 1 \right)^3} = 1\,\,,\,\,{\left( 2 \right)^3} = 8\,\,,\,\,{\left( 3 \right)^3} = 27\,\,,{\left( 4 \right)^3} = 64\,\,,\,\,{\left( 5 \right)^3} = 125\,\,\,$
Therefore,
Sum $ = \,{\left( 1 \right)^3} + {\left( 2 \right)^3} + {\left( 3 \right)^3} + {\left( 4 \right)^3} + {\left( 5 \right)^3}$
Sum= $1 + 8 + 27 + 64 + 125$
Sum= $225$
Now, using the formula
$sum\, = \,{\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2}$
Put, $n = 5$
$sum\, = \,{\left[ {\dfrac{{5\left( {5 + 1} \right)}}{2}} \right]^2}$
$ \Rightarrow {\left[ {\dfrac{{5 \times 6}}{2}} \right]^2}$
$ \Rightarrow {\left[ {5 \times 3} \right]^2}$
$ \Rightarrow {\left[ {15} \right]^2}$
$ \Rightarrow 225$
Hence, verified

Now, to find the sum of the cubes of the first ten natural numbers.
Put, $n = 10$
$sum\, = \,{\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2}$
$ \Rightarrow {\left[ {\dfrac{{10\left( {10 + 1} \right)}}{2}} \right]^2}$
$ \Rightarrow {\left[ {\dfrac{{10 \times 11}}{2}} \right]^2}$
$ \Rightarrow {\left[ {5 \times 11} \right]^2}$
$ \Rightarrow {\left[ {55} \right]^2}$
$ \Rightarrow 3025$
Hence, the sum of the cube of the first ten natural numbers is $3025$.

Note: The given formula is true for only natural numbers. The given formula is very important in cases when the value of n is very large. You can’t add all the numbers individually. Like the cube, the formula for the sum of first n natural numbers and the sum of the square of first n natural numbers also exist.