Binomial-Theorem-And-Its-Simple-Applications Question 234
Question: For all $ n\in N, $ $ 1+\frac{1}{1+2}+\frac{1}{1+2+3}+…+\frac{1}{1+2+3+…+n} $ is equal to
Options:
A) $ \frac{3n}{n+1} $
B) $ \frac{n}{n+1} $
C) $ \frac{2n}{n-1} $
D) $ \frac{2n}{n+1} $
Correct Answer: DShow Answer
Answer:
Solution:
$ +\frac{1}{1+2+3+…+n}=\frac{2n}{n+1} $
i.e. $ P(n):1+\frac{1}{1+2}+\frac{1}{1+2+3}+…+\frac{2}{n(n+1)}=\frac{2n}{n+1} $
Step I: For $ n=1, $ $ P(1):1=\frac{2\times 1}{1+1}=\frac{2}{2}=1, $ which is true.
Step II: Let it is true for n = k,
i.e., $ 1+\frac{1}{1+2}+\frac{1}{1+2+3}+…+\frac{2}{k(k+1)}=\frac{2k}{k+1} $ ..(i)
Step III: For $ n=k+1, $
$ ( 1+\frac{1}{1+2}+\frac{1}{1+2+3}+…+\frac{2}{k(k+1)} )+\frac{2}{(k+1)(k+2)} $
$ =\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)} $ [Using equation (i)]
$ =\frac{2k(k+2)+2}{(k+1)(k+2)}=\frac{2[ k^{2}+2k+1 ]}{(k+1)(k+2)} $
[Taking 2 common in numerator part]
$ =\frac{2{{(k+1)}^{2}}}{(k+1)(k+2)}=\frac{2(k+1)}{k+2}=\frac{2(k+1)}{(k+1)+1} $
Therefore, $ P(k+1) $ is true, when P (k) is true, hence, from the principle of mathematical induction, the statement is true for all natural numbers n.
BETA
