If the sum `3/1^2+5/(1^2+2^2)+7/(1^2+2^2+3^2)+. . . +` upto 20 terms is equal to `k/21,` then k is equal to

Options

A. 120
B. 180
C. 240
D. 60

Answer

To solve the problem, we need to evaluate the sum:

\[ S = \sum_{n=1}^{20} \frac{2n + 1}{\sum_{k=1}^{n} k^2} \]

**Step 1: Identify the general term.**

The general term of the series is given by:

\[ T_n = \frac{2n + 1}{\sum_{k=1}^{n} k^2} \]

The formula for the sum of the squares of the first \( n \) natural numbers is:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

Thus, we can rewrite \( T_n \):

\[ T_n = \frac{2n + 1}{\frac{n(n + 1)(2n + 1)}{6}} = \frac{6(2n + 1)}{n(n + 1)(2n + 1)} \]

**Step 2: Simplify the term.**

Notice that \( 2n + 1 \) cancels out:

\[ T_n = \frac{6}{n(n + 1)} \]

**Step 3: Rewrite \( T_n \) using partial fractions.**

We can express \( \frac{6}{n(n + 1)} \) using partial fractions:

\[ \frac{6}{n(n + 1)} = 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]

**Step 4: Write the sum \( S \).**

Now, we can write the sum \( S \):

\[ S = \sum_{n=1}^{20} T_n = \sum_{n=1}^{20} 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]

**Step 5: Evaluate the telescoping series.**

The series is telescoping:

\[ S = 6 \left( \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{20} - \frac{1}{21} \right) \right) \]

Most terms cancel out, and we are left with:

\[ S = 6 \left( 1 - \frac{1}{21} \right) = 6 \left( \frac{20}{21} \right) = \frac{120}{21} \]

**Step 6: Find \( k \).**

According to the problem, we have \( S = \frac{k}{21} \). Thus, we can equate:

\[ \frac{120}{21} = \frac{k}{21} \]

This implies:

\[ k = 120 \]

**Final Answer:**

The value of \( k \) is \( \boxed{120} \).

---