Question

1³+2³+3³+.... +100³=5050² then 1+2+3+......+100=?

Answer 1

294.35 lol just guessing

Answer 2

You can use the sum of series formula for an AP. Sum of n terms of an AP is given by:
$s = \frac{n}{2} (2a + (n - 1)d)$
And a is the first term, d is the common difference.

In this case a=1 and d=1.

So,

$s = \frac{n(n + 1)}{2}$

$s = \frac{100 \times 101}{2} = 5050$

Or

You may know,

$\Sigma_{k=1}^{n} k = \frac{n(n+1)}{2}$

And

$\Sigma_{k=1}^{n} k^{3}=\frac{n^{2}(n+1)^{2}}{4}$

You can observe that the sum of first n numbers cubed equals to the square of sum of first n numbers.

Thus,

$1 + 2 + 3 + 4 + 5....+ 100 = \sqrt{ {1}^{3} + {2}^{3} + {3}^{3} + {4}^{3} + {5}^{3}..... + { 100}^{3} } = \sqrt{ {5050}^{2} } \\ =5050$