Question

Prove that the sum of all the numbers of the form $n^3$ which lie between 100 and 100000 is 53261.

Answer 1

The first number will be $5^3=125$
Now${20}^3=400\times 20=8000,{21}^3=441\times 21=9261$
It is evident that ${22}^3$ will be greater than 10,000.
Hence we have to find
$5^3+6^3+7^3+.....+{21}^3$
$=(1^3+2^3+.....+{21}^3)-(1^3+2^3+3^3+4^3)$
$=\sum n^{3}forn=21-\sum k^{3}fork=4$
$\left[{\frac{n(n+1)}{2}}^2\right]-\left[{\frac{k(k+1)}{2}}^2\right]=(22\times 11{)}^2-(2\times 5{)}^2$
$=(231-10)(231+10)=221\times 241=53261$