Question

How do you use sigma notation to find the sum for $\dfrac{1}{{1.3}} + \dfrac{1}{{2.4}} + \dfrac{1}{{3.5}} + ..... + \dfrac{1}{{10.12}}$ ?

Answer 1

Hint : Here first we need to observe the pattern in the question. Is there any pattern? If yes then try to write that pattern in a general way. Then we will get the number of terms or the value of n upto which the pattern is given and then we will write it in the form of sigma or what we say is summation.

Complete step by step solution:
Given that,
$\dfrac{1}{{1.3}} + \dfrac{1}{{2.4}} + \dfrac{1}{{3.5}} + ..... + \dfrac{1}{{10.12}}$
Observing the first term we can write it as $\dfrac{1}{{1.\left( {1 + 2} \right)}}$
Next second term can be written as $\dfrac{1}{{2.\left( {2 + 2} \right)}}$
Third term can be written as $\dfrac{1}{{3.\left( {3 + 2} \right)}}$
And lats term is $\dfrac{1}{{10.\left( {10 + 2} \right)}}$
So the given pattern can be written as
$= \dfrac{1}{{1.\left( {1 + 2} \right)}} + \dfrac{1}{{2.\left( {2 + 2} \right)}} + \dfrac{1}{{3.\left( {3 + 2} \right)}} + ..... + \dfrac{1}{{10.\left( {10 + 2} \right)}}$
This can be observed as $\dfrac{1}{{n\left( {n + 2} \right)}}$ for n equals to 1,2,3….10
So the sigma can be written as,
$= \sum\limits_{n = 1}^{10} {\dfrac{1}{{n\left( {n + 2} \right)}}}$
This is the correct answer.
So, the correct answer is “ $\sum\limits_{n = 1}^{10} {\dfrac{1}{{n\left( {n + 2} \right)}}}$ ”.

Note : Here note that the terms are summed up with a pattern. The number of terms that are to be summed up are counted on the basis of last term in the pattern. It has n equals to 10 and the n+2 becomes 12. So we have taken n from 1 to 10 and not zero. Sigma is nothing but the summation notation.