Sequences And Series Question 3
Question 3 - 24 January - Shift 2
If $\frac{1^{3}+2^{3}+3^{3}+\ldots \ldots \text{. upto } n \text{ terms }}{1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \ldots \text{. upto } n \text{ terms }}=\frac{9}{5}$, then the value of $n$ is ________
Show Answer
Answer: 5
Solution:
Formula: Sum of first $n$ natural numbers, Sum of squares of first $n$ natural numbers, Sum of cubes of first $n$ natural numbers,
$1^{3}+2^{3}+3^{3} \ldots . .+n^{3}=(\frac{n(n+1)}{2})^{2}$
$1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \ldots+n terms = \sum _{r=1}^{n} r(2 r+1)=\sum _{r=1}^{n}(2 r^{2}+r)$
$=\frac{2 \cdot n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}$
$=\frac{n(n+1)}{6}(2(2 n+1)+3)$
$=\frac{n(n+1)}{2} \times \frac{(4 n+5)}{3}$
$=\frac{\frac{n^{2}(n+1)^{2}}{4}}{\frac{n(n+1)}{2} \times \frac{(4 n+5)}{3}}=\frac{9}{5}$
$\Rightarrow \frac{5 n(n+1)}{2}=\frac{9(4 n+5)}{3}$
$\Rightarrow 15 n(n+1)=18(4 n+5)$
$\Rightarrow 15 n^{2}+15 n=72 n+90$
$\Rightarrow 15 n^{2}-57 n-90=0 \Rightarrow 5 n^{2}-19 n-30=0$
$\Rightarrow(n-5)(5 n+6)=0$
$\Rightarrow n=\frac{-6}{5}$ or 5
$\Rightarrow n=5$.
