\( \dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+....\) is the given series.
The nth term of the series is
\( a_n=\dfrac{1}{n(n+1)}\)
By partial fractions the above equation can be written as
\(a_n=\dfrac{1}{n}-\dfrac{1}{n+1}\)
Therefore,
\( a_1=\dfrac{1}{1}-\dfrac{1}{2}\)
\( a_2=\dfrac{1}{2}-\dfrac{1}{3}\)
\( a_3=\dfrac{1}{3}-\dfrac{1}{4}\)
\( a_n=\dfrac{1}{n}-\dfrac{1}{n+1}\)
Add the above terms.
\( a_1+a_2+a_3+.......+a_n\)
= \([ \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+.....\dfrac{1}{n}]\)
= \( [ \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+.....\dfrac{1}{n+1}]\)
The sum of n terms of a series is given by the equation
\( S_n= \displaystyle\sum_{k=1}^{n}a_k\)
The sum of n terms of the given series is
\( S_n= 1-\dfrac{1}{n-1}\)
= \( \dfrac{n+1-1}{n+1}=\dfrac{n}{n-1}\)
Therefore, the sum of n terms of the series is \( =\dfrac{n}{n-1}\)
Answered by Pragya Singh | 1 year agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.
Find the geometric means of the following pairs of numbers:
(i) 2 and 8
(ii) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).