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Question: Consider the following expression and solve it accordingly: \[\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfra...

Consider the following expression and solve it accordingly:
12.5+15.8+18.11+...........1(3n−1).(3n+2)\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...........\dfrac{1}{\left( 3n-1 \right).\left( 3n+2 \right)}

Explanation

Solution

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(n)S\left( n \right)by some expressionn(k)n(k). Then we have to prove that the given expression also holds good for n=k+1n=k+1.

Complete step by step answer:
Let S(n)=12.5+15.8+18.11+...........1(3n−1).(3n+2)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(n)S\left( n \right)for n=1
L.H.S==12.5+15.8+18.11+...........1(3(1)−1).(3(1)+2)=\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)}
=12.5+15.8+18.11+...........12.5=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...........\dfrac{1}{2.5}
L. H. S=12.5=110=\dfrac{1}{2.5}=\dfrac{1}{10}. . . . . . . . . . . . . . . . . . . . (1)
R. H. S=12.5=110=\dfrac{1}{2.5}=\dfrac{1}{10}. . . . . . . . . . . . . . . . . . . . (2)
So the given expression is true for n=1.
Step-2
Assume that S(n)S\left( n \right)is true for some n(k)n(k)
12.5+15.8+18.11+...........1(3k−1).(3k+2)=k6k+4\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(n)S\left( n \right)for n=k+1n=k+1
L. H. S=12.5+15.8+18.11+...........1(3(k+1)−1).(3(k+1)+2)=\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)}
=k2(3k+2)+1(3k+2).(3k+5)=\dfrac{k}{2(3k+2)}+\dfrac{1}{\left( 3k+2 \right).\left( 3k+5 \right)}. . . . . . . . . . . . . . . . . . . . . . . (4)

& =\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{aligned}$$ So the given expression is also true for $$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