A) \[\frac{n}{2}({{n}^{2}}-1)\]
B) \[\frac{n}{2}({{n}^{2}}+1)\]
C) \[\frac{2}{n}({{n}^{2}}+1)\]
D) \[\frac{2}{n}({{n}^{2}}-1)\]
Correct Answer: B
Solution :
\[{{T}_{n}}=\frac{1}{2}({{n}^{2}}-n+2),\] this is the first term of two \[{{n}^{th}}\] row. Also, the terms of \[{{n}^{th}}\]row form an A.P. with its first term as \[\frac{1}{2}({{n}^{2}}-n+2)\] and a common difference of 1. Hence the sum of the terms in the \[{{n}^{th}}\] row is \[p:q:r=(-1\pm \sqrt{3})K:2K:(-1\mp \sqrt{3})K\].
You need to login to perform this action.
You will be redirected in
3 sec
