Question

If $\sum _{r=1}^n{T}_r=\frac{n}{8}(n+1)(n+2)(n+3)$ and $\sum _{r=1}^n\frac{1}{T_r}=\frac{n^2+3n}{4\sum _{k=1}^{p}k}$, then $p$ is equal to

• A. $n+1$
• B. $n$
• C. $n-1$
• D. $2n$

Answer 1

The correct option is A $n+1$

Given : $\sum _{r=1}^n{T}_r=\frac{n}{8}(n+1)(n+2)(n+3)$
So,
$S_n=\frac{n}{8}(n+1)(n+2)(n+3)\Rightarrow S_n=\frac{1}{2}\left[\frac{n(n+1)(n+2)(n+3)}{4}\right]\Rightarrow T_n=\frac{1}{2}\left[n\right(n+1\left)\right(n+2\left)\right]$​​

Now,
$\frac{1}{T_n}=\frac{2}{n(n+1)(n+2)}\Rightarrow \frac{1}{T_n}=\frac{(n+2)-n}{n(n+1)(n+2)}\Rightarrow \frac{1}{T_n}=\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\Rightarrow \sum _{r=1}^n\frac{1}{T_r}=\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}+\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}⋮+\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\Rightarrow \frac{n^2+3n}{4\sum _{k=1}^{p}k}=\frac{1}{2}-\frac{1}{(n+1)(n+2)}\Rightarrow \frac{n^2+3n}{4\sum _{k=1}^{p}k}=\frac{n^2+3n}{2(n+1)(n+2)}\Rightarrow 4\sum _{k=1}^{p}k=2(n+1)(n+2)\Rightarrow \sum _{k=1}^{p}k=\frac{(n+1)(n+2)}{2}\therefore p=n+1$