Question

Find $1^3+2^3+3^3+4^3+5^3+........+n^3$

• A. ${\left[\frac{n\left(n+1\right)}{2}\right]}^2$
• B. ${\left[\frac{n\left(2n+1\right)}{2}\right]}^2$
  
• C. ${\left[\frac{n\left(n+1\right)}{4}\right]}^2$
• D. ${\left[\frac{n\left(n+1\right)}{6}\right]}^2$

Answer 1

Answer: (A)

${\left[\frac{n\left(n+1\right)}{2}\right]}^2$

${\sum }_{r=1}^n{r}^4-{(r-1)}^4=n^4-(n-1{)}^4(n-1{)}^4-(n-2{)}^4+...+3^4-2^4+2^4-1^4+1^4-0^4=n^4$

But $r^4-{(r-1)}^4=r^4-(r^4-4r^3+6r^2-4r+1)=4r^3-6r^2+4r-1$

$\therefore \sum 4r^3-6r^2+4r-1$

$=4\sum r^3-6\sum r^2+4\sum r-\sum 1$

$=4\sum r^3-6\frac{n(n+1)(2n+1)}{6}+4\frac{n(n+1)}{2}-n$

$=4\sum r^3-n(n+1)(2n+1)+2n(n+1)-n=n^4$

$4\sum r^3=n^4+n(n+1)(2n+1)-2n(n+1)+n$

$=n^4+2n^3+3n^2+n-2n^2-2n+n$

$=n^2(n^2+2n+1)$

$=n^2(n+1{)}^2$

$\therefore \sum r^3=\frac{n^2(n+1{)}^2}{4}={\left(\frac{n(n+1)}{2}\right)}^2$