If the sum of the series (1)/(1 (1+d))+(1)/((1+d)(1+2d))+ +(1)/((1+9d)(1+10d)) is equal to…

Question

If the sum of the series $\frac{1}{1\cdot (1+d)}+\frac{1}{(1+d)(1+2d)}+\ldots +\frac{1}{(1+9d)(1+10d)}$ is equal to 5, then $50d$ is equal to :

Accepted Answer

$\frac{1}{1\cdot (1+d)}+\frac{1}{(1+d)(1+2d)}+\ldots +\frac{1}{(1+9d)(1+10d)}=5$

Multiply and divide by $d$

$\begin{gathered} \frac{1}{d}[\frac{d}{1	imes (1+d)}+\frac{d}{(1+d)(1+2d)}+\ldots +\frac{1}{(1+9d)(1+10d)}]=5 \ \frac{1}{d}[(1-\frac{1}{1+d})+(\frac{1}{1+d}-\frac{1}{1+2d})+\ldots +(\frac{1}{1+9d}-\frac{1}{1+10d})]=5 \ \frac{1}{d}[1-\frac{1}{1+10d}]=5 \ \frac{1}{d}[\frac{1+10d-1}{1+10d}]=5 \end{gathered}$

$\begin{gathered} \frac{10}{1+10d}=5 \ 1+10d=2 \ d=\frac{1}{10} \ 50d=50	imes \frac{1}{10}=5 \end{gathered}$

Solutions · 1