Let k = \frac{1}{{1 \times 2 \times 3}} + \frac{2}{{3 \times 4 \times 5}} + \frac{3}{{5 \times 6 \times 7}} + .... +  upto infinite terms, then k is

1/2

1/3

1/4

1/5

To find the value of \( k \), given by the infinite series:

\[
k = \frac{1}{{1 \times 2 \times 3}} + \frac{2}{{3 \times 4 \times 5}} + \frac{3}{{5 \times 6 \times 7}} + \ldots
\]

we need to find a general formula for the \(n\)-th term and then sum the series.

### Step 1: General Formula for the \(n\)-th Term

Let's denote the \(n\)-th term of the series as \( a_n \). We observe a pattern in the numerators and denominators:

\[
a_n = \frac{n}{{(2n-1)(2n)(2n+1)}}
\]

### Step 2: Simplify the \(n\)-th Term

We can simplify \( a_n \) using partial fraction decomposition. We want to express it in the form:

\[
\frac{n}{{(2n-1)(2n)(2n+1)}} = \frac{A}{2n-1} + \frac{B}{2n} + \frac{C}{2n+1}
\]

Multiplying through by \((2n-1)(2n)(2n+1)\), we get:

\[
n = A(2n)(2n+1) + B(2n-1)(2n+1) + C(2n-1)(2n)
\]

Expand and collect terms involving powers of \(n\):

\[
n = (4A + 4B + 4C) n^2 + (2A + 2B - 2C) n + (-B)
\]

To match the left-hand side, the coefficients must satisfy:

1. \(4A + 4B + 4C = 0\)
2. \(2A + 2B - 2C = 1\)
3. \(-B = 0\)

From \(-B = 0\), we have \(B = 0\).

Now, substitute \(B = 0\) into the other equations:

1. \(4A + 4C = 0 \implies A = -C\)
2. \(2A - 2C = 1 \implies 2(-C) - 2C = 1 \implies -4C = 1 \implies C = -\frac{1}{4}\)

Thus, \(A = \frac{1}{4}\), \(B = 0\), and \(C = -\frac{1}{4}\).

### Step 3: Rewrite the \(n\)-th Term Using Partial Fractions

Substitute the values of \(A\), \(B\), and \(C\):

\[
a_n = \frac{\frac{1}{4}}{2n-1} + \frac{0}{2n} + \frac{-\frac{1}{4}}{2n+1} = \frac{1}{4(2n-1)} - \frac{1}{4(2n+1)}
\]

### Step 4: Form the Series and Recognize a Telescoping Pattern

Now, \(k\) becomes:

\[
k = \sum_{n=1}^\infty \left( \frac{1}{4(2n-1)} - \frac{1}{4(2n+1)} \right)
\]

This is a telescoping series, where most terms cancel out. The series simplifies to:

\[
k = \frac{1}{4} \left( \frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \ldots \right)
\]

All intermediate terms cancel, and we are left with:

\[
k = \frac{1}{4} \times 1 = \frac{1}{4}
\]

### Conclusion

The value of \(k\) is:

\[
\boxed{\frac{1}{4}}
\]

Would you like more details or have any questions?

### Related Questions

1. How do we derive partial fractions for more complex fractions?
2. What is a telescoping series, and why does it simplify summations?
3. Can we apply similar methods to series involving different patterns?
4. How can we determine the convergence of an infinite series?
5. How would the value of \(k\) change if the series started with a different term?

**Tip:** Always look for patterns like telescoping in infinite series, as they can simplify the problem significantly.

Explore the solution to the infinite series problem to find the value of k using telescoping series technique. Learn how to derive the general formula for the nth term, simplify using partial fractions, and sum the series. Suitable for college-level mathematics.

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