Question

Find the sum of 17 consecutive cubical numbers starting from 3.​

Answer 1

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Answer 2

Given,

A series of 17 consecutive cubical numbers starting from 3.

To Find,

Sum of a series of 17 consecutive cubical numbers starting from 3.

Solution,

The formula for sum n consecutive Cubes = $(\frac{n(n+1)}{2})^{2}$

The sum of 19 consecutive cubical numbers = $(\frac{19(19+1)}{2})^{2}$ = $(\frac{19(20)}{2})^{2}$ = $36100$

The sum of 2 consecutive cubical numbers = $(\frac{2(2+1)}{2})^{2}$ = $9$

Thus the sum of 17 consecutive cubical numbers starting from 3 is = $36100 - 9 = 36091$

Henceforth, the answer is 36091.