Sequence And Series Question 125
Question: The sum of the series $ 3.6+4.7+5.8+…….. $ upto $ (n-2) $ terms
[EAMCET 1980]
Options:
A) $ n^{3}+n^{2}+n+2 $
B) $ \frac{1}{6}(2n^{3}+12n^{2}+10n-84) $
C) $ n^{3}+n^{2}+n $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ S=3\ .\ 6+4\ .\ 7+……. $ upto $ n-2 $ terms $ =(1\ .\ 4+2\ .\ 5+3\ .\ 6+4\ .\ 7+……… $ upto $ n $ terms) - 14 $ =\Sigma n(n+3)-14=\frac{1}{6}(2n^{3}+12n^{2}+10n)-14 $ $ =( \frac{2n^{3}+12n^{2}+10n-84}{6} ),\ $ where $ n=3,\ 4,\ 5……. $ Trick: $ S_1=18,\ S_2=46 $ Now put in options $ (n-2)=1,\ 2\ \ i.e.\ \ n=3,\ 4 $ Obviously (b) gives the values.
BETA
