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Question: Evaluate \(1.2 + 2.3 + 3.4 + ....... + n(n + 1) = \dfrac{n}{3}(n + 1)(n + 2)\)....

Evaluate 1.2+2.3+3.4+.......+n(n+1)=n3(n+1)(n+2)1.2 + 2.3 + 3.4 + ....... + n(n + 1) = \dfrac{n}{3}(n + 1)(n + 2).

Explanation

Solution

We know that ∑x=1nx=n(n+1)2 , ∑x=1nx2=n(n+1)(2n+1)6\sum\limits_{x = 1}^n {x = \dfrac{{n(n + 1)}}{2}}\,,\,\sum\limits_{x = 1}^n {{x^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}.So we are given 1.2+2.3+3.4+.......+n(n+1)1.2 + 2.3 + 3.4 + ....... + n(n + 1).So its general form is n(n+1)n(n + 1).And we need to find ∑n=1nn(n+1)\sum\limits_{n = 1}^n n (n + 1) that is ∑n=1nn2+∑n=1nn\sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n n .Using these formulas and prove that 1.2+2.3+3.4+.......+n(n+1)=n3(n+1)(n+2)1.2 + 2.3 + 3.4 + ....... + n(n + 1) = \dfrac{n}{3}(n + 1)(n + 2).

Complete step-by-step answer:
Here we are given that
1.2+2.3+3.4+.......+n(n+1)1.2 + 2.3 + 3.4 + ....... + n(n + 1)
So we can write this as
∑n=1nn(n+1)\sum\limits_{n = 1}^n n (n + 1)
So upon expanding by putting n=1,2,3,4,.......,nn = 1,2,3,4,.......,n, we get
1(1+1)+2(2+1),,,,,,,,,,+n(n+1)1(1 + 1) + 2(2 + 1),,,,,,,,,, + n(n + 1)
1.2+2.3+3.4+..........+n(n+1)1.2 + 2.3 + 3.4 + .......... + n(n + 1)
So we get the value we need to find, so we can write that
1.2+2.3+3.4+.......+n(n+1)1.2 + 2.3 + 3.4 + ....... + n(n + 1) as ∑n=1nn(n+1)\sum\limits_{n = 1}^n n (n + 1)
Now opening the bracket by multiplying the terms
∑n=1nn2+∑n=1nn\sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n n
So from (1) and (2), we get that
∑i=1ni=n(n+1)2\sum\limits_{i = 1}^n {i = \dfrac{{n(n + 1)}}{2}} and ∑i=1ni2=n(n+1)(2n+1)6\sum\limits_{i = 1}^n {{i^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}
So we can expand by the above formula, we get that
So ∑n=1nn2+∑n=1nn\sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n n
=n(n+1)(2n+1)6+n(n+1)2= \dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{{n(n + 1)}}{2}
As ∑x=1nx=n(n+1)2\sum\limits_{x = 1}^n {x = \dfrac{{n(n + 1)}}{2}}
∑x=1nx2=n(n+1)(2n+1)6\sum\limits_{x = 1}^n {{x^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}
Now upon simplification, we can take n(n+1)2\dfrac{{n(n + 1)}}{2} common
We get
=n(n+1)2(2n+13+1)= \dfrac{{n(n + 1)}}{2}\left( {\dfrac{{2n + 1}}{3} + 1} \right)
=n(n+1)2(2n+1+33)= \dfrac{{n(n + 1)}}{2}\left( {\dfrac{{2n + 1 + 3}}{3}} \right)
=n(n+1)2(2n+43)= \dfrac{{n(n + 1)}}{2}\left( {\dfrac{{2n + 4}}{3}} \right)
Now taking 22 common from 2n+42n + 4, we get
=2n(n+1)(n+2)6= \dfrac{{2n(n + 1)(n + 2)}}{6}
=n(n+1)(n+2)3= \dfrac{{n(n + 1)(n + 2)}}{3}
Hence proved.

Note: We know that when 1+2+3+4+.......+n1 + 2 + 3 + 4 + ....... + n is given, then its value is equal to n(n+1)2\dfrac{{n(n + 1)}}{2}.This can be proved by using AP as we know that 1,2,3,4,.........,n1,2,3,4,.........,n are in AP with the common difference d=1d=1 and first term a=1a=1.So the sum of nth term is given by =n2(2a+(n−1)d) = \dfrac{n}{2}\left( {2a + (n - 1)d} \right),Here a=1,d=1a = 1,d = 1
Sum=n2(2+(n−1)) = \dfrac{n}{2}\left( {2 + (n - 1)} \right) =n(n+1)2 = \dfrac{{n(n + 1)}}{2}
Students should remember these formulas for solving the questions.
∑i=1ni=1+2+3+4+........+n=n(n+1)2\sum\limits_{i = 1}^n {i = 1 + 2 + 3 + 4 + ........ + n = \dfrac{{n(n + 1)}}{2}}
∑i=1ni2=12+22+32+.......+n2=n(n+1)(2n+1)6\sum\limits_{i = 1}^n {{i^2}} = {1^2} + {2^2} + {3^2} + ....... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}
∑i=1ni3=13+23+33+.......+n3=(n(n+1)2)2\sum\limits_{i = 1}^n {{i^3}} = {1^3} + {2^3} + {3^3} + ....... + {n^3} = {\left( {\dfrac{{n(n + 1)}}{2}} \right)^2}.