Sequence And Series Question 145
Question: The sum of the series $ 1.3.5+.2.5.8+3.7.11+……… $ upto $ ’n’ $ terms is
[Dhanbad Engg. 1972]
Options:
A) $ \frac{n,(n+1)(9n^{2}+23n+13)}{6} $
B) $ \frac{n,(n-1)(9n^{2}+23n+12)}{6} $
C) $ \frac{(n+1)(9n^{2}+23n+13)}{6} $
D) $ \frac{n,(9n^{2}+23n+13)}{6} $
Show Answer
Answer:
Correct Answer: A
Solution:
Given series is, $ 1\ .\ 3\ .\ 5+2\ .\ 5\ .\ 8+3\ .\ 7\ .\ 11+…….+n(2n+1)(3n+2)\ $ So, $ T_{n}=n(2n+1)(3n+2)=n,[6n^{2}+4n+3n+2] $ $ T_{n}=6n^{3}+7n^{2}+2n $ Now, sum $ =6\Sigma n^{3}+7\Sigma n^{2}+2\Sigma n $ $ =6{{[ \frac{1}{2}n(n+1) ]}^{2}}+7[ \frac{1}{6}n(n+1)(2n+1) ]+2[ \frac{1}{2}n(n+1) ] $ $ =\frac{1}{6}n(n+1)[9n,(n+1)+7(2n+1)+6] $ $ =\frac{1}{6}n,(n+1)[9n^{2}+9n+14n+7+6] $ $ =\frac{n,(n+1)(9n^{2}+23n+13)}{6} $ .
BETA
