Question

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

Answer 1

Hint: 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 \[a = 3\], \[S = \sum\limits_{k = 1}^n {{K^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} \].