Given:
The sum of terms
[tex]4^3+5^3+6^3..........+13^3[/tex]
Required:
Find sum.
Explanation:
We know sum of cube of first n terms of natural numbers
[tex]\sum_{n\mathop{=}1}^{\infty}n^3=[\frac{n(n+1)}{2}]^2[/tex]
Now,
[tex]\begin{gathered} =(1^3+2^3+....+13^3)-(1^3+2^3+3^3) \\ =[\frac{13(13+1)}{2}]^2-36 \\ =8281-36 \\ =8245 \end{gathered}[/tex]
Answer:
The sum of terms is 8245.
