Question

Prove $ 2 + 5 + 8 + 11 + .... + (3n - 1) = \dfrac{1}{2}n(3n + 1)$,$n \in N $ .

Answer 1

Hint: An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. The expression given in the question is in the form of Arithmetic Progression. Initially, we will find the common difference ‘d’ value. It can be found by subtracting the first term of the given expression from the second term. Then, we will check for the first term ‘a’. Finally, we will apply the formula of Sum of ‘n’ terms in A.P, which is given by ${S_n}$.

Formula Used:
 $ {S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] $

Complete step-by-step answer:
As per the given question, it is given that:
L.H.S = $ 2 + 5 + 8 + 11 + .... + (3n - 1) $
R.H.S = $ \dfrac{1}{2}n(3n + 1) $
From L.H.S part,
First term = $ a = 2 $--- (1)
The product of the difference between second and first term is given by:
$ d = {d_{\text{next term}}} - {d_{\text{current term}}} $ --- (2)
So, by using equation (2), we can find the common difference value:
$ {d_1} = 5 - 2 = 3 $,
$ {d_2} = 8 - 5 = 3 $,
\[{d_3} = 11 - 8 = 3\]
Therefore, the common difference is:
$ d = 3 $--- (3)
Sum of ‘n’ terms in Arithmetic Progression is given by:
$ {S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] $ --- (4)
Now, substitute equation (1) and (3) in equation (4),
 $ {S_n} = \dfrac{n}{2}\left[ {2(2) + (n - 1)3} \right] \to {S_n} = \dfrac{n}{2}\left[ {4 + 3n - 3} \right] $
$ {S_n} = \dfrac{n}{2}\left[ {3n + 1} \right] \to {S_n} = \dfrac{1}{2}n\left[ {3n + 1} \right] $ --- (5)
From equation (5), it has been proved that:
L.H.S = R.H.S
$ 2 + 5 + 8 + 11 + .... + (3n - 1) = \dfrac{1}{2}n(3n + 1) $

Note: After looking at the question, the given expression is in the form of Arithmetic Progression. It can be found by different techniques either by finding the sum of N terms in A.P or by the method of mathematical induction.
also by putting different values of n checking RHS and LHS
For any given ‘n’ terms equation in the form of Arithmetic Progression, Sum of ‘n’ terms can be found by
 $ {S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] $