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Question: The sum of nth term of the series \(1.2.5 + 2.3.6 + 3.4.7 + ......n\) terms is \(\dfrac{1}{{12}}n\le...

The sum of nth term of the series 1.2.5+2.3.6+3.4.7+......n1.2.5 + 2.3.6 + 3.4.7 + ......n terms is 112n(n+1)(3n2+23n+34)\dfrac{1}{{12}}n\left( {n + 1} \right)\left( {3{n^2} + 23n + 34} \right)
A) True
B) False

Explanation

Solution

1.2.5+2.3.6+3.4.7+......n1.2.5 + 2.3.6 + 3.4.7 + ......n. First, we have to find the general term. For this we can see 1.2.5+2.3.6+3.4.7+......n1.2.5 + 2.3.6 + 3.4.7 + ......n varies according to n.(n+1).(n+4)n.\left( {n + 1} \right).\left( {n + 4} \right) hence general term is n.(n+1)(n+4)n.\left( {n + 1} \right)\left( {n + 4} \right) on submission we know ∑n.(n+1)(n+4)\sum {n.\left( {n + 1} \right)\left( {n + 4} \right)} we know that ∑n3=n2(n+1)24\sum {{n^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} ,∑n2=n(n+1)(2n+1)6\sum {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} and ∑n=n(n+1)2\sum {n = \dfrac{{n\left( {n + 1} \right)}}{2}}

Complete step by step solution:
1.2.5+2.3.6+3.4.7+......n1.2.5 + 2.3.6 + 3.4.7 + ......n terms
1.(1+1).(1+4)+2.(2+1).(2+4)+3.(3+1).(3+4)+.......n1.\left( {1 + 1} \right).\left( {1 + 4} \right) + 2.\left( {2 + 1} \right).\left( {2 + 4} \right) + 3.\left( {3 + 1} \right).\left( {3 + 4} \right) + .......n
Therefore, the general term
⇒n.(n+1)(n+4)\Rightarrow n.\left( {n + 1} \right)\left( {n + 4} \right)
⇒n3+5n2+4n\Rightarrow {n^3} + 5{n^2} + 4n
Now, sum of the series will be
∑n.(n+1)(n+4)\sum {n.\left( {n + 1} \right)\left( {n + 4} \right)}
∑n3+5n2+4n\sum {{n^3} + 5{n^2} + 4n}
⇒∑n3+5∑n2+4∑n\Rightarrow \sum {{n^3} + } 5\sum {{n^2} + 4} \sum n
We know that ∑n3=n2(n+1)24\sum {{n^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} ,∑n2=n(n+1)(2n+1)6\sum {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} and ∑n=n(n+1)2\sum {n = \dfrac{{n\left( {n + 1} \right)}}{2}}
⇒n2(n+1)24+5n(n+1)(2n+1)6+4n(n+1)2\Rightarrow \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + \dfrac{{5n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \dfrac{{4n\left( {n + 1} \right)}}{2}
Taking common n(n+1)n\left( {n + 1} \right)
⇒n(n+1)2(n(n+1)2+5(2n+1)3+4)\Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{n\left( {n + 1} \right)}}{2} + \dfrac{{5\left( {2n + 1} \right)}}{3} + 4} \right)
⇒n(n+1)2(3n2+3n+20n+10+246)\Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{3{n^2} + 3n + 20n + 10 + 24}}{6}} \right)
⇒112n(n+1)(3n2+23n+34)\Rightarrow \dfrac{1}{{12}}n\left( {n + 1} \right)\left( {3{n^2} + 23n + 34} \right)

Note:
In these type of question 1st we find the general term then apply rule submission ∑n3=n2(n+1)24\sum {{n^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} ,∑n2=n(n+1)(2n+1)6\sum {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} and ∑n=n(n+1)2\sum {n = \dfrac{{n\left( {n + 1} \right)}}{2}} for constant xn where x is constant.