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 $$\sum _{k=1}^n{K}^3={\displaystyle \frac{n^2{\left(n+1\right)}^2}{4}}$$, 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 $$S=\sum _{k=1}^n{K}^3=1^3+2^3+3^3+...+n^3$$ gives the sum of the cube of $$n^{th}$$ powers of the first $$n$$ positive integers.
The general formula to compute the is:
$$S=\sum _{k=1}^n{K}^3={\displaystyle \frac{n^2{\left(n+1\right)}^2}{4}}$$ -------(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
$$\Rightarrow S={\displaystyle \frac{{15}^2{\left(15+1\right)}^2}{4}}$$
$$\Rightarrow S={\displaystyle \frac{{15}^2{\left(16\right)}^2}{4}}$$
As we know square numbers i.e., $${15}^2=225$$ and $${16}^2=256$$, then
$$\Rightarrow S={\displaystyle \frac{225\left(256\right)}{4}}$$
$$\Rightarrow S={\displaystyle \frac{57600}{4}}$$
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 $$S=\sum _{k=1}^n{K}^a=1^a+2^a+3^a+...+n^a$$ 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 $$a=1$$, $$S=\sum _{k=1}^{n}K={\displaystyle \frac{n\left(n+1\right)}{2}}$$
If $$a=2$$, $$S=\sum _{k=1}^n{K}^2={\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{6}}$$
If $$a=3$$, $$S=\sum _{k=1}^n{K}^3={\displaystyle \frac{n^2{\left(n+1\right)}^2}{4}}$$.