Find the sum of:
 
73+83+93+...+243
 
Answer:
 
We know that, the sum of the cubes of first \(n\) natural numbers \(=\) ii+iii.
 
73+83+93+...+243 \(=\) \((1^3 + 2^3 + 3^3 + ... 24^3)\) \(-\) 13+23+33+43+53+63
 
\(=\) \(-\)
 
\(=\)