Question

If $1^3+2^3+3^3+.........+m^3=3025,$ then $m=?$

• A. 15
• B. 45
• C. 14
• D. 10

Answer 1

Answer: (D)

10

Given

$1^3+2^3+3^3+......+m^3=3025{\left(\frac{m(m+1)}{2}\right)}^2=3025$

$\Rightarrow m^2{(m+1)}^2=3025\times 4\Rightarrow m(m+1)=\pm 110$
$\Rightarrow m^2+m\mp 110=0$
$\therefore m=\frac{-1\pm \sqrt{1\pm 4\cdot 1\cdot 110}}{2}m=\frac{-1\pm \sqrt{1+440}}{2}=\frac{-1\pm 21}{2}$

we don not consider ${}^{\prime }{-}^{\prime }$ sign.Since the discriminant will be less than zero and will not be defined.

$m=10$