Question

# Prove by the principle of mathematical induction that for all $n \in N$: $1 + 4 + 7 + .... + (3n - 2) = \dfrac{1}{2}n(3n - 1)$

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Hint: Here in this question the concept of mathematical induction will get used which states that if the statement is true for n=k then it will also be true for its successor i.e. n=k+1. This is known as principle of mathematical induction

Let the given statement be named as P(n) so it can be written as$P(n) = 1 + 4 + 7 + .... + (3n - 2) = \dfrac{1}{2}n(3n - 1)$, $n \in N$
For n=1
$\Rightarrow P(1) = \dfrac{1}{2}(1)(3 - 1) = 1$ which is true
Now we will assume that P(m) is true for some positive integers ‘m’$\Rightarrow P(m) = 1 + 4 + 7 + .... + (3m - 2) = \dfrac{1}{2}m(3m - 1)$
Now according to principle of mathematical induction if a statement is true for n=m then it will also be true for its successor i.e. n=m+1, therefore we shall prove that P(m+1) is also true.$\Rightarrow P(m + 1) = 1 + 4 + 7 + .... + (3(m + 1) - 2) = \dfrac{1}{2}(m + 1)((3m + 3) - 1)$$\Rightarrow 1 + 4 + 7 + .... + (3m + 3 - 2) = \dfrac{1}{2}(m + 1)(3m + 2)$ $\Rightarrow 1 + 4 + 7 + .... + (3m + 1) = \dfrac{1}{2}(m + 1)(3m + 2)$
L.H.S=$1 + 4 + 7 + .... + (3m + 1)$
R.H.S=$\dfrac{1}{2}(m + 1)(3m + 2)$
Now we will prove L.H.S=R.H.S
$\Rightarrow 1 + 4 + 7 + .... + (3m - 2) + (3m + 1)$ (Adding term before 3m+1)
Here we will apply formula of sum of arithmetic progression i.e. ${S_n} = \dfrac{n}{2}(2a + (n - 1)d)$ $\Rightarrow {S_n} = \dfrac{m}{2}(2(1) + (m - 1)(4 - 1))$ (Here a=1, d=4-1 and m is the total number of terms)
Now putting and solving the equation we will get a sum of arithmetic progression. $\Rightarrow {S_n} = \dfrac{m}{2}(2 + (m - 1)3)$
$\Rightarrow {S_n} = \dfrac{m}{2}(2 + 3m - 3) = \dfrac{m}{2}(3m - 1)$
$\Rightarrow \dfrac{{m(3m - 1)}}{2} + (3m + 1)$
Now taking L.C.M we will get
$\Rightarrow \dfrac{{m(3m - 1) + 2(3m + 1)}}{2}$
$\Rightarrow \dfrac{{3{m^2} - m + 6m + 2}}{2}$
$\Rightarrow \dfrac{{3{m^2} + 5m + 2}}{2} = \dfrac{{3{m^2} + 3m + 2m + 2}}{2}$
(Splitting the equation to find the factors of quadratic equation)
$\Rightarrow \dfrac{{3m(m + 1) + 2(m + 1)}}{2}$
$\Rightarrow \dfrac{{(3m + 2)(m + 1)}}{2}$=R.H.S
Therefore P(m+1) holds true.
Thus, by principle of mathematical induction, for all $n \in N$, P(n) holds true.

Note: Some students may find difficulty in splitting the quadratic equation whose coefficient of degree two variable is not one so below is the explanation of it: -
Let’s take above equation as an example: -$3{m^2} + 5m + 2$
Now comparing with general quadratic equation $a{x^2} + bx + c$
Here a=3, b=5, c=2
Step1. Multiply ab i.e. $ab = 3 \times 2 = 6$
Step2. Now split 6 into that factors so that its sum is equal to ‘b’
$\Rightarrow 3 \times 2 = 6$ Factors are 3 and 2 whose sum is 5.
Therefore 5m will get split into 3m and 2m and the equation will become $3{m^2} + 3m + 2m + 2$