Find the sum of the series.
$1^{3} + 2^{3} + 3^{3} + . . . . + 20^{3}$

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Solution

We know $1^{3} + 2^{3} + 3^{3} + . . . . + 20^{3} = 20 \sum 1 n^{3}$
$= {(\frac{20 (20 + 1)}{2})}^{2}$ using $n \sum k = 1 k^{3} = {[\frac{n (n + 1)}{2}]}^{2}$ .
$= {(\frac{20 \times 21}{2})}^{2} = (210)^{2} = 44100$ .

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Find the sum of the series.13+23+33+....+203

We know 13+23+33+....+203=20∑1n3=(20(20+1)2)2 using n∑k=1k3=[n(n+1)2]2.=(20×212)2=(210)2=44100.

Find the sum of the series.
$1^{3} + 2^{3} + 3^{3} + . . . . + 20^{3}$

We know $1^{3} + 2^{3} + 3^{3} + . . . . + 20^{3} = 20 \sum 1 n^{3}$
$= {(\frac{20 (20 + 1)}{2})}^{2}$ using $n \sum k = 1 k^{3} = {[\frac{n (n + 1)}{2}]}^{2}$ .
$= {(\frac{20 \times 21}{2})}^{2} = (210)^{2} = 44100$ .