Question

If ${\sum }_{r=1}^n{T}_r=\frac{n}{8}(n+1)(n+2)(n+3)$ then find ${\sum }_{r=1}^n\frac{1}{T_r}$

• A. $\frac{n^2+3n}{2(n+1)(n+2)}$
• B. $\frac{n^2+3n}{2(n+1)(n+2)(n+3)}$
• C. $\frac{n^2-3n}{2(n+1)(n+2)}$
• D. None of these

Answer 1

The correct option is A $\frac{n^2+3n}{2(n+1)(n+2)}$

$\because T_n=S_n-S_{n-1}$
$={\sum }_{r=1}^n{T}_r-{\sum }_{r=1}^{n-1}{T}_r$
$=\frac{n(n+1)(n+2)(n+3)}{8}-\frac{(n-1)n(n+1)(n+2)}{8}$
$=\frac{n(n+1)(n+2)}{8}\left[\right(n+3)-(n-1\left)\right]$
$=\frac{n(n+1)(n+2)}{8}\left(4\right)$
$=\frac{n(n+1)(n+2)}{2}$
$\Rightarrow \frac{1}{T_n}=\frac{2}{n(n+1)(n+2)}\cdots \left(1\right)$
$\therefore V_n=\frac{2}{(n+1)(n+2)}$
then, $V_{n-1}=\frac{2}{n(n+1)}$
$\therefore V_n-V_{n-1}=\frac{2}{(n+1)(n+2)}-\frac{2}{n(n+1)}$
$=-\frac{4}{n(n+1)(n+2)}$
$=-2\frac{1}{T_n}$ from $\left(1\right)$
$\therefore \frac{1}{T_n}=-\frac{1}{2}(V_n-V_{n-1})$
Putting $n=1,2,3,...n$, we get
$\frac{1}{T_1}=-\frac{1}{2}(V_1-V_0)$
$\frac{1}{T_2}=-\frac{1}{2}(V_2-V_1)$
$\frac{1}{T_3}=-\frac{1}{2}(V_3-V_2)$
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$\frac{1}{T_n}=-\frac{1}{2}(V_n-V_{n-1})$
$\therefore \frac{1}{T_1}+\frac{1}{T_2}+\frac{1}{T_3}+....+\frac{1}{T_n}=-\frac{1}{2}(V_n-V_0)$
$\Rightarrow {\sum }_{r=1}^n\frac{1}{T_r}=-\frac{1}{2}\left[\frac{2}{(n+1)(n+2)}-\frac{2}{2}\right]$
$=\frac{(n+1)(n+2)-2}{2(n+1)(n+2)}$
$=\frac{n^2+3n}{2(n+1)(n+2)}$