Question

(3) Find the sum of the cubes of the first 7 natural numbers.​

Answer 1

This is the answer for your question

Answer 2

Answer:

Formula :

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

Now,

first 7 natural numbers.

${1}^{3} + {2}^{3} + {3}^{3} + ... + {6}^{3} + {7}^{3} \\ \: so \: \: \: \: n = 7 \\ \\ now \\ \\ sum \: = \: \{ { \frac{7 \times (7 + 1)}{2} } \}^{2} \\ \: \: \: \: \: \: \: \: = ( { \frac{7 \times 8}{2} })^{2} \\ \: \: \: \: \: \: \: \: \: = ( { \frac{56}{2} })^{2} \\ \: \: \: \: \: \: \: \: \: = {(28)}^{2} \\ \: \: \: \: \: \: \: \: \: = \: 784$