Binomial Theorem And Its Simple Applications Question 140
Question: The sum of n terms of $ 1+\frac{1+2}{2}+\frac{1+2+3}{3}+… $ is
Options:
A) $ \frac{m(n+3)}{4} $
B) $ \frac{(n+3)}{3} $
C) $ \frac{n(n-3)}{3} $
D) $ \frac{n(n-3)}{4} $
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ n^{th}termorT_{n}=\frac{1+2+3+…+n}{n}, $
$ S_{n}=\sum{T_{n}=\sum{\frac{1+2+…+n}{n}=\sum{\frac{n(n+1)}{2n}}}} $
$ \Rightarrow \sum{\frac{n+1}{2}=\frac{1}{2}(\sum{n}+\sum{1})=\frac{1}{2}[ \frac{n(n+1)}{2}+n ]} $
$ =\frac{n(n+1)+2n}{4}=\frac{n(n+3)}{4} $
BETA
