Question

1³+2³+3³+...200³ find sum of series

Answer 1

$\sf{\red{\large{\underline{Question:-}}}}$

1³+2³+3³+...200³ find sum of series

$\sf{\red{\large{\underline{Answer:-}}}}$

$\rm{\blue{\underline{Formula:}}}$

1³ + 2³ + 3³ +.......+ n³ = $\bf{\green{\underline{\boxed{[{\frac{n(n\:+\:1)}{2}]}^{2}}}}}$

$\sf{\pink{\underline{Here,}}}$

We have to find 1³ + 2³ + 3³ +.........+200³

Here, n = 200

$\rm{So\ finding\ sum\ of\ cubes: }$

Sum of cubes = ${\sf{[\frac{200(200\:+\:1)}{2}]}^{2}}$

⤇${\sf{[\frac{200\:*201}{2}]}^{2}}$

⤇(100 * 201)²

⤇ (20100)²

⤇40,40,10,000

$\bf{\underline{\boxed{\red{\large{Answer\:=\:40,40,10,000}}}}}$