Question

If ${\sum }_{r=1}^n{r}^3-{\sum }_{p=1}^n{\sum }_{m=1}^p{\sum }_{r=1}^{m}1=80$, then possible of $n$ can be -

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is B 4

${\sum }_{r=1}^n{r}^3-{\sum }_{p=1}^n{\sum }_{m=1}^p{\sum }_{r=1}^{m}1=80$

Taking L.H.S

Using summation forms

$=\frac{n^2\cdot (n+1{)}^2}{4}-{\sum }_{p=1}^n{\sum }_{m=1}^{p}m$

$=\frac{n^2\cdot (n+1{)}^2}{4}-{\sum }_{p=1}^n\frac{p(p+1)}{2}$

$=\frac{n^2\cdot (n+1{)}^2}{4}-\frac{1}{2}{\sum }_{p=1}^n(p^2+p)$

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

Further solving

Try hit and trial method from the given option

$\therefore$ By hit and trial n=4