Calculating the Sum of the Series

We are asked to find the sum to 'n' terms of the series:

\(\dfrac{1}{1.2.3}+\dfrac{3}{2.3.4}+\dfrac{5}{3.4.5}+\dfrac{7}{4.5.6}+...\)

Let the k-th term of the series be denoted by \(T_k\).

Finding the General Term \(T_k\)

Let's observe the pattern in the terms:

• The numerator is 1, 3, 5, 7, ... This is an arithmetic progression with the first term 1 and common difference 2. The k-th term of this arithmetic progression is \(a_k = 1 + (k-1)2 = 1 + 2k - 2 = 2k-1\).
• The denominator is a product of three consecutive integers. For the 1st term, it is 1.2.3. For the 2nd term, it is 2.3.4. For the k-th term, it is \(k(k+1)(k+2)\).

Thus, the general term of the series is:

\(T_k = \dfrac{2k-1}{k(k+1)(k+2)}\)

Partial Fraction Decomposition of \(T_k\)

To find the sum, we decompose the general term \(T_k\) using partial fractions. We look for a form like:

\(T_k = \dfrac{A}{k} + \dfrac{B}{k+1} + \dfrac{C}{k+2}\)

Multiplying both sides by \(k(k+1)(k+2)\), we get:

\(2k-1 = A(k+1)(k+2) + B k(k+2) + C k(k+1)\)

We can find the constants A, B, and C by substituting specific values of k:

• Setting \(k=0\): \(-1 = A(1)(2) + B(0) + C(0) \implies -1 = 2A \implies A = -1/2\)
• Setting \(k=-1\): \(2(-1)-1 = A(0) + B(-1)(-1+2) + C(0) \implies -3 = B(-1)(1) \implies -3 = -B \implies B = 3\)
• Setting \(k=-2\): \(2(-2)-1 = A(0) + B(0) + C(-2)(-2+1) \implies -5 = C(-2)(-1) \implies -5 = 2C \implies C = -5/2\)

So, the partial fraction decomposition is:

\(T_k = \dfrac{-1/2}{k} + \dfrac{3}{k+1} + \dfrac{-5/2}{k+2}\)

\(T_k = -\dfrac{1}{2k} + \dfrac{3}{k+1} - \dfrac{5}{2(k+2)}\)

Let's rewrite \(T_k\) to facilitate telescoping sum by grouping terms:

\(T_k = \dfrac{1}{2} \left( -\dfrac{1}{k} + \dfrac{6}{k+1} - \dfrac{5}{k+2} \right)\)

We can try to express this in the form \(f(k) - f(k+1)\). Let's try \(f(k) = \dfrac{P}{k} + \dfrac{Q}{k+1}\). Then \(f(k+1) = \dfrac{P}{k+1} + \dfrac{Q}{k+2}\).

\(f(k) - f(k+1) = \left(\dfrac{P}{k} + \dfrac{Q}{k+1}\right) - \left(\dfrac{P}{k+1} + \dfrac{Q}{k+2}\right) = \dfrac{P}{k} + \dfrac{Q-P}{k+1} - \dfrac{Q}{k+2}\)

We want \(T_k = \dfrac{1}{2} [f(k) - f(k+1)]\). So we need to match the coefficients:

\(\dfrac{P}{2} = -\dfrac{1}{2} \implies P = -1\)

\(\dfrac{Q-P}{2} = 3 \implies Q-P = 6 \implies Q - (-1) = 6 \implies Q+1 = 6 \implies Q = 5\)

\(\dfrac{-Q}{2} = -\dfrac{5}{2} \implies -Q = -5 \implies Q = 5\)

The values of P and Q match, so we have successfully written \(T_k\) in the telescoping form.

Let \(f(k) = -\dfrac{1}{k} + \dfrac{5}{k+1}\). Then \(f(k+1) = -\dfrac{1}{k+1} + \dfrac{5}{k+2}\).

And \(T_k = \dfrac{1}{2} [f(k) - f(k+1)]\).

Calculating the Sum \(S_n\) using Telescoping Series

The sum of the first n terms is \(S_n = \sum_{k=1}^{n} T_k\).

\(S_n = \sum_{k=1}^{n} \dfrac{1}{2} [f(k) - f(k+1)]\)

\(S_n = \dfrac{1}{2} \sum_{k=1}^{n} [f(k) - f(k+1)]\)

The sum \(\sum_{k=1}^{n} [f(k) - f(k+1)]\) is a telescoping sum:

\(\sum_{k=1}^{n} [f(k) - f(k+1)] = (f(1) - f(2)) + (f(2) - f(3)) + ... + (f(n) - f(n+1))\)

\(= f(1) - f(n+1)\)

Now we calculate \(f(1)\) and \(f(n+1)\):

\(f(1) = -\dfrac{1}{1} + \dfrac{5}{1+1} = -1 + \dfrac{5}{2} = \dfrac{-2+5}{2} = \dfrac{3}{2}\)

\(f(n+1) = -\dfrac{1}{n+1} + \dfrac{5}{(n+1)+1} = -\dfrac{1}{n+1} + \dfrac{5}{n+2}\)

Combine \(f(n+1)\) terms:

\(f(n+1) = \dfrac{-1(n+2) + 5(n+1)}{(n+1)(n+2)} = \dfrac{-n-2 + 5n+5}{(n+1)(n+2)} = \dfrac{4n+3}{(n+1)(n+2)}\)

Now substitute these into the sum formula \(S_n = \dfrac{1}{2} [f(1) - f(n+1)]\):

\(S_n = \dfrac{1}{2} \left[ \dfrac{3}{2} - \dfrac{4n+3}{(n+1)(n+2)} \right]\)

Find a common denominator inside the brackets:

\(S_n = \dfrac{1}{2} \left[ \dfrac{3(n+1)(n+2) - 2(4n+3)}{2(n+1)(n+2)} \right]\)

\(S_n = \dfrac{3(n^2 + 2n + n + 2) - (8n + 6)}{4(n+1)(n+2)}\)

\(S_n = \dfrac{3(n^2 + 3n + 2) - 8n - 6}{4(n+1)(n+2)}\)

\(S_n = \dfrac{3n^2 + 9n + 6 - 8n - 6}{4(n+1)(n+2)}\)

\(S_n = \dfrac{3n^2 + n}{4(n+1)(n+2)}\)

Factor out n from the numerator:

\(S_n = \dfrac{n(3n+1)}{4(n+1)(n+2)}\)

Comparing with Options

The calculated sum to 'n' terms is \(\dfrac{n(3n+1)}{4(n+1)(n+2)}\), which matches Option 1.

Revision Table: Key Steps for Sum of Series

Additional Information: Telescoping Series and Partial Fractions

A telescoping series is a series where most of the terms cancel out, leaving only a few terms. The sum of a telescoping series \(\sum_{k=1}^n (f(k) - f(k+1))\) is \(f(1) - f(n+1)\).

Partial fraction decomposition is a technique used to break down a rational function (a fraction where the numerator and denominator are polynomials) into a sum of simpler fractions. This is often useful for integration or for finding the sum of series where the general term is a rational function.

For a rational function like \(\dfrac{P(x)}{(x-a)(x-b)(x-c)}\), the decomposition might be \(\dfrac{A}{x-a} + \dfrac{B}{x-b} + \dfrac{C}{x-c}\). In our series problem, the denominator had factors \(k\), \(k+1\), and \(k+2\).

Successfully expressing the general term \(T_k\) as a difference \(f(k) - f(k+1)\) is the key step in summing this type of series using the telescoping method.