Question

The sum of series ${1^3} + {2^3} + {3^3} + ... + {15^3}$ is
A.22000
B.10000
C.14400
D.15000

Answer 1

Here in this question given a series of cubes of the first 15 natural numbers. we have to find their sum. For this we use the formula sum of cubes of n natural numbers is $\sum\limits_{k = 1}^n {{K^3}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$, where n is the positive integer and it indicates number of terms in series. On substituting the n value in formula and further simplify by using a basic arithmetic operation to get the required solution.

Complete answer: Before solving the problem, we will discuss the formula of sum of the cubes of first n natural numbers.
The series $S = \sum\limits_{k = 1}^n {{K^3}} = {1^3} + {2^3} + {3^3} + ... + {n^3}$ gives the sum of the cube of ${n^{th}}$ powers of the first $n$ positive integers.
The general formula to compute the is:
$S = \sum\limits_{k = 1}^n {{K^3}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$ -------(1)
Now consider the given series:
$S = {1^3} + {2^3} + {3^3} + ... + {15^3}$
We have to find the sum of the given cubic series.
The given series had 15 terms
$\therefore \,\,\,n = 15$
On substituting $n$ value in the formula of sum of cube numbers, then
$\Rightarrow \,\,\,S = \dfrac{{{{15}^2}{{\left( {15 + 1} \right)}^2}}}{4}$
$\Rightarrow \,\,\,S = \dfrac{{{{15}^2}{{\left( {16} \right)}^2}}}{4}$
As we know square numbers i.e., ${15^2} = 225$ and ${16^2} = 256$, then
$\Rightarrow \,\,\,S = \dfrac{{225\left( {256} \right)}}{4}$
$\Rightarrow \,\,\,S = \dfrac{{57600}}{4}$
On simplification, we get
$\therefore \,\,\,S = 14,400$
Hence, the sum of the cubes of the first 15 natural numbers is 14,400.
Therefore, option (3) is the correct answer.

Note:
To find the sum we have to know some formulas:
If the series $S = \sum\limits_{k = 1}^n {{K^a}} = {1^a} + {2^a} + {3^a} + ... + {n^a}$ gives the sum of the ${a^{th}}$ power of the first $n$ positive numbers, where $a$ and $n$ are positive integers.
The generalized formulas to compute the sum of first few values of $a$ are as follows:
If $a = 1$, $S = \sum\limits_{k = 1}^n {K = \dfrac{{n\left( {n + 1} \right)}}{2}}$
If $a = 2$, $S = \sum\limits_{k = 1}^n {{K^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}$
If $a = 3$, $S = \sum\limits_{k = 1}^n {{K^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}}$.