If 1^3+2^3+3^3+ up to n terms13+25+37+ up to n terms = 95, then the value of n is ________…

Question

If $\dfrac{1^3+2^3+3^3+\cdots 	ext{ up to } n 	ext{ terms}}{1\cdot3+2\cdot5+3\cdot7+\cdots 	ext{ up to } n 	ext{ terms}} = \dfrac{9}{5}$, then the value of $n$ is \_\_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

Numerator $=\left(\dfrac{n(n+1)}{2}\right)^2$. Denominator: $\sum_{r=1}^n r(2r+1) = \dfrac{n(n+1)(4n+5)}{6}$ (combining $2\sum r^2+\sum r$).

$$\frac{n^2(n+1)^2/4}{n(n+1)(4n+5)/6} = \frac{3n(n+1)}{2(4n+5)} = \frac95.$$

$$15n(n+1) = 18(4n+5) \implies 15n^2-57n-90=0 \implies 5n^2-19n-30=0 \implies (n-5)(5n+6)=0.$$

Since $n$ is a positive integer, $n=5$.

Solutions · 1