If the sum of the series frac{1}{1 cdot(1+d)} | vedclass

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(B) The given series is $S = \sum_{k=0}^{9} \frac{1}{(1+kd)(1+(k+1)d)} = 5$. Using the method of partial fractions,we can write each term as: $\frac{1

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$5$

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(B) The given series is $S = \sum_{k=0}^{9} \frac{1}{(1+kd)(1+(k+1)d)} = 5$. Using the method of partial fractions,we can write each term as: $\frac{1}{(1+kd)(1+(k+1)d)} = \frac{1}{d} \left( \frac{1}{1+kd} - \frac{1}{1+(k+1)d} \right)$. Summing from $k=0$ to $9$: $S = \frac{1}{d} \left[ \left( \frac{1}{1} - \frac{1}{1+d} \right) + \left( \frac{1}{1+d} - \frac{1}{1+2d} \right) + \dots + \left( \frac{1}{1+9d} - \frac{1}{1+10d} \right) \right] = 5$. This is a telescoping series,so it simplifies to: $\frac{1}{d} \left( 1 - \frac{1}{1+10d} \right) = 5$. $\frac{1}{d} \left( \frac{1+10d-1}{1+10d} \right) = 5$. $\frac{1}{d} \left( \frac{10d}{1+10d} \right) = 5$. $\frac{10}{1+10d} = 5$. $10 = 5(1+10d) \implies 10 = 5 + 50d$. $50d = 5$.

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

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