If the sum of the series 1/(1·(1+d)) + 1/((1+d)(1+2d)) + ... + 1/((1+9d)(1+10d)) is equal to 5, then

Q: Why did we multiply and divide by $d$? Isn't that changing the value of the expression?

Great observation! We aren't changing the value because we are multiplying by $\frac{d}{d}$, which is effectively multiplying by 1. We do this to create a 'bridge' that allows us to split the fraction into a difference of two simpler terms.

Q: How do I know for sure that it's a telescoping series?

A series is telescoping if the general term can be written as $f(r) - f(r+1)$. When you sum these, the $-f(2)$ cancels with $+f(2)$, $-f(3)$ with $+f(3)$, and so on. If you see a pattern where the denominator factors are in arithmetic progression, it is almost certainly a telescoping series.

Q: What if $d=0$? Does the formula still hold?

If $d=0$, the denominators become $1 \cdot 1$, $1 \cdot 1$, etc. The series would just be $1+1+1...$ which would diverge. In this problem, $d$ must be non-zero for the expression to be well-defined and equal to 5.

Q: I got lost in the algebra. How do I handle the last term correctly?

Always write out the first two terms and the last term explicitly. The first term is $\frac{1}{d}(1 - \frac{1}{1+d})$ and the last term is $\frac{1}{d}(\frac{1}{1+9d} - \frac{1}{1+10d})$. When you add them, everything in the middle vanishes, leaving only the very first and very last pieces.

Q: Is there a shortcut to finding the sum of a telescoping series?

Yes! For a series of the form $\sum \frac{1}{(a+(r-1)d)(a+rd)}$, the sum is always $\frac{1}{d} (\text{First term's start} - \text{Last term's end})$. It's a powerful tool, but always derive it once to ensure you understand the mechanics!

Question

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 :

Options:

(A) 10

(B) 5

(C) 15

(D) 20

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given series is:

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

At first glance, this appears to be a daunting collection of fractions. However, in JEE Advanced mathematics, such complexity is often a mask for hidden symmetry. We will use the method of **telescoping series** to simplify this expression.

### The Anatomy of the Term

To solve this, we focus on the general term $T_r$. The $r$-th term of the series is given by:

$$
T_r = \frac{1}{(1+(r-1)d)(1+rd)}
$$

Notice that the second factor of the denominator, $(1+rd)$, is exactly the first factor of the next term, $T_{r+1}$. This structural alignment is the hallmark of a **telescoping series**.

### The Art of Splitting

We must decompose the fraction by examining the difference between the two factors in the denominator:

$$
(1+rd) - (1+(r-1)d) = d
$$

This constant difference $d$ is our "golden key." By multiplying and dividing the general term by $d$, we can rewrite the numerator as the difference of the denominator factors:

$$
T_r = \frac{1}{d} \left[ \frac{(1+rd) - (1+(r-1)d)}{(1+(r-1)d)(1+rd)} \right]
$$

Distributing the denominator allows us to transform the product into a simple subtraction:

$$
T_r = \frac{1}{d} \left[ \frac{1}{1+(r-1)d} - \frac{1}{1+rd} \right]
$$

### The Great Collapse

Now, we evaluate the sum $S = \sum_{r=1}^{10} T_r$. Expanding this summation reveals the cancellation pattern:

$$
S = \frac{1}{d} \left[ \left(1 - \frac{1}{1+d}\right) + \left(\frac{1}{1+d} - \frac{1}{1+2d}\right) + \dots + \left(\frac{1}{1+9d} - \frac{1}{1+10d}\right) \right]
$$

Every intermediate term is annihilated by its neighbor. Like a telescope collapsing into itself, only the first and last components remain:

$$
S = \frac{1}{d} \left[ 1 - \frac{1}{1+10d} \right] = \frac{1}{d} \left[ \frac{1+10d-1}{1+10d} \right] = \frac{10}{1+10d}
$$

### Final Calculation

We are given that the sum equals 5. Therefore, we set up the following equation:

$$
\frac{10}{1+10d} = 5
$$

Solving for $d$, we find $1+10d = 2$, which simplifies to $10d = 1$. The problem asks for the value of $50d$. Multiplying both sides by 5, we obtain:

**$50d = 5$**

Animated Solution for Mathematics - Sequence and Series: If the sum of the series $\frac{1}{1 \cdot ( 1 + d )} + \frac{1}{( 1 + d ) ( 1 + 2 d )} + \dots + \frac{1}{( 1 + 9 d ) ( 1 + 10 d )}$ is equal to 5, then $50 d$ is equal to :

Select Answer:

Visualized Solution

Observe the Series Structure

Identify the General Term $T_{r}$

The Method of Differences

Splitting into Partial Fractions

Expanding the Summation

The Telescoping Cancellation

Simplified Expression for Sum

Equating to the Given Value

Final Calculation of $50 d$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Anatomy of the Term

The Art of Splitting

The Great Collapse

Final Calculation

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