Question

the sum of the square of the first n natural numbers is 285 while the sum of the cubes is 2025 find the value of n

Answer 1

Answer:

The value of n is 9

Step-by-step explanation:

The sum of the square of the first n natural numbers is 285 while the sum of the cubes is 2025 find the value of n

$\text{Formula used:}$

$1^2+2^2+...........+n^2=\frac{n(n+1)(2n+1)}{6}$

$1^3+2^3+...........+n^3=[\frac{n(n+1)}{2}]^2$

$\text{Given:}$

$1^2+2^2+...........+n^2=285$

$\frac{n(n+1)(2n+1)}{6}=285$......(1)

$\text{Also,}$

$1^3+2^3+...........+n^3=2025$

$\implies\:[\frac{n(n+1)}{2}]^2=2025$

$\implies\:\frac{n(n+1)}{2}=\sqrt{2025}$

$\implies\:\frac{n(n+1)}{2}=45$

$\implies\:n^2+n-90=0$

$\implies\:(n+10)(n-9)=0$

$\implies\:n=-10,9$

$\text{But n cannot be negative}$

$\therefore\:\bf{n=9}$