Question

find the sum of the consecutive cube upto 45 cube

Answer 1

formula:
sum of consecutive cubes upto 'n'=
$({ \frac{n(n + 1)}{2} })^{2}$
so,sum upto 45³
={(45)(46)/2}²
={(45)(23)}²
=(1035)²
=10,71,225.
hope it'll help
mark it as brainliest..

Answer 2

Answer:

The sum of the consecutive cube up to 45 cube is 1,071,225 .

Step by step explanation:

Hey Mate,

The sum of the consecutive cube = 1³ + 2³ + 3³ + 4³ + 5³ ....... + n³.

=>  $\frac{(n)^{2}(n+1)^{2}}{4}$

So, The sum of consecutive cube up to 45 cube is

1³ + 2³ + 3³ + 4³ + 5³ ....... + 45³

=> $\frac{(45)^{2}(45+1)^{2}}{4}$

=> $\frac{(4284900)}{4}$

=> 1,071,225

The sum of the consecutive cube up to 45 cube is 1,071,225 .