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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

**Complete step-by-step answer:**

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 $3{m^2} + 3m + 2m + 2$

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