Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

Consider the following expression and solve it accordingly:
\[\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...........\dfrac{1}{\left( 3n-1 \right).\left( 3n+2 \right)}\]

Answer
VerifiedVerified
569.1k+ views
Hint: First consider the given expression as some function or represent by a particular notation. Then calculate the value of the given expression in both left hand side and right hand side for n=1. Now assume the total sum \[S\left( n \right)\]by some expression\[n(k)\]. Then we have to prove that the given expression also holds good for \[n=k+1\].

Complete step by step answer:
Let \[S\left( n \right)=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...........\dfrac{1}{\left( 3n-1 \right).\left( 3n+2 \right)}\]
Step-1
Prove \[S\left( n \right)\]for n=1
L.H.S=\[=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...........\dfrac{1}{\left( 3\left( 1 \right)-1 \right).\left( 3\left( 1 \right)+2 \right)}\]
\[=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...........\dfrac{1}{2.5}\]
L. H. S\[=\dfrac{1}{2.5}=\dfrac{1}{10}\]. . . . . . . . . . . . . . . . . . . . (1)
R. H. S\[=\dfrac{1}{2.5}=\dfrac{1}{10}\]. . . . . . . . . . . . . . . . . . . . (2)
So the given expression is true for n=1.
Step-2
Assume that \[S\left( n \right)\]is true for some \[n(k)\]
\[\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...........\dfrac{1}{\left( 3k-1 \right).\left( 3k+2 \right)}=\dfrac{k}{6k+4}\]. . . . . . . . . . . . . . . . . . (3)
Step-3
Prove \[S\left( n \right)\]for \[n=k+1\]
L. H. S\[=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...........\dfrac{1}{\left( 3\left( k+1 \right)-1 \right).\left( 3\left( k+1 \right)+2 \right)}\]
\[=\dfrac{k}{2(3k+2)}+\dfrac{1}{\left( 3k+2 \right).\left( 3k+5 \right)}\]. . . . . . . . . . . . . . . . . . . . . . . (4)
\[\begin{align}
  & =\dfrac{3{{k}^{2}}+5k+2}{2\left( 3k+2 \right).\left( 3k+5 \right)} \\
 & =\dfrac{\left( 3k+2 \right)\left( k+1 \right)}{2\left( 3k+2 \right).\left( 3k+5 \right)} \\
 & =\dfrac{k+1}{6k+10} \\
\end{align}\]
So the given expression is also true for \[n=k+1\]

Note:
For this type of problem we have to use the principle of mathematical induction to solve the given expression. The principle of mathematical induction states that if S be a conjecture involving a positive integer n. S is true for all positive integers if the following conditions hold good.
S is true for n=1
If S is true for n=k
6. Then S should also be true for n=k+1