Question

Find the sum of the series \[\dfrac{1}{3\times 6}+\dfrac{1}{6\times 9}+\dfrac{1}{9\times 12}+........100\text{ }terms\]
1. \[\dfrac{25}{160}\]
2. \[\dfrac{1}{6}\]
3. \[\dfrac{100}{909}\]
4. \[\dfrac{25}{152}\]

Answer 1

Hint: Here to solve questions like this you will first try and see if the sequence given to us is in arithmetic progression or geometric progression. If it’s not, try to see if parts of the terms of the sequence are in either geometric or arithmetic progression. Once we know that all we need to do is find the nth term of the series which gives us an idea about the sequence in question and then solve the sequence to find the nth term here and solve the question of finding the hundredth term.

Complete step-by-step solution:
The sequence given to us is
\[S=\dfrac{1}{3\times 6}+\dfrac{1}{6\times 9}+\dfrac{1}{9\times 12}+........\]
Now we can see that S in itself is neither an arithmetic progression or geometric progression therefore we see the different parts of each and every term in this sequence. Now checking that we can see that
Here \[3,6,9....\] is in an arithmetic progression with \[a=3;d=3\]
Now to find the nth term here we use the formula
\[{{t}_{n}}=a+(n-1)d\]
Which gives
\[{{t}_{n}}=3n\]
Now for the second part of each term we can see that \[6,9,12....\] is in an arithmetic progression with \[a=6;d=3\]
Now the nth term according to formula used above will be
\[{{t}_{n}}=6+(n-1)3=3n+3\]
Now hence we can see that
\[{{S}_{n}}=\dfrac{1}{3\times 6}+\dfrac{1}{6\times 9}+\dfrac{1}{9\times 12}+........\dfrac{1}{(3n)(3n+3)}\]
Now since we can see that the common difference everywhere is three therefore we multiply and divide the complete sequence by three
\[{{S}_{n}}=\dfrac{3}{3}\left( \dfrac{1}{3\times 6}+\dfrac{1}{6\times 9}+\dfrac{1}{9\times 12}+........\dfrac{1}{(3n)(3n+3)} \right)\]
Now taking the numerator inside
\[{{S}_{n}}=\dfrac{1}{3}\left( \dfrac{3}{3\times 6}+\dfrac{3}{6\times 9}+\dfrac{3}{9\times 12}+........\dfrac{3}{(3n)(3n+3)} \right)\]
Now we can write the numerators as
\[{{S}_{n}}=\dfrac{1}{3}\left( \dfrac{6-3}{3\times 6}+\dfrac{9-6}{6\times 9}+\dfrac{12-9}{9\times 12}+........\dfrac{(3n+3)-(3n)}{(3n)(3n+3)} \right)\]
Now separating each term we get
\[{{S}_{n}}=\dfrac{1}{3}\left( \dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+........\dfrac{1}{3n}-\dfrac{1}{3n+3} \right)\]
Now cancelling all the same terms we are left with
\[{{S}_{n}}=\dfrac{1}{3}\left( \dfrac{1}{3}-\dfrac{1}{3n+3} \right)\]
Taking the LCM of denominator we get
\[{{S}_{n}}=\dfrac{1}{3}\left( \dfrac{3n+3-3}{3(3n+3)} \right)\]
Solving
\[{{S}_{n}}=\dfrac{n}{3(3n+3)}\]
Now here in this question there are \[100\] terms therefore
\[{{S}_{n}}=\dfrac{100}{3(3(100)+3)}\]
\[{{S}_{n}}=\dfrac{100}{909}\]
Therefore Option 3, $\dfrac{100}{909}$ is the correct option.

Note: Now in such types of questions at a common place students get stuck in understanding which is the geometric or arithmetic progression. So always take care when it comes to that step. Sometimes in the problem we are directly not given A.P and G.P but the terms follow a pattern where we simplify that pattern. For this either we can break the given terms into a difference of two terms as we did in the solution part or can write the terms in a general form having a variable and then solve by using sigma method or Vn method. Also students must know the basic formulas of sequence and series to be able to solve this question.