Sequence And Series Question 137
Question: The sum of all the products of the first $ n $ natural numbers taken two at a time is
Options:
A) $ \frac{1}{24}n(n-1)(n+1)(3n+2) $
B) $ \frac{n^{2}}{48}(n-1)(n-2) $
C) $ \frac{1}{6}n(n+1)(n+2)(n+5) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
We know that $ {{{ \frac{n}{2}(n+1) }}^{2}}={{(1+2+……+n)}^{2}}=\sum\limits_1^{n}{r^{2}}+2\sum\limits_{s<t}{st} $
$ \Rightarrow $ $ \sum\limits_{s<t}{st}=\frac{1}{2}{ \frac{n^{2}{{(n+1)}^{2}}}{4}-\frac{n(n+1)(2n+1)}{6} } $ $ =\frac{n}{24}(n-1)(n+1)(3n+2) $ . Trick: $ S_{n}=1\ .\ 2+2\ .\ 3+3\ .\ 4+……..+(n-1)\ .\ n $ Check by putting $ (n-1)=1,\ 2\ i.e.,\ n=2,\ 3 $ in the options.
BETA
