A) \[\frac{3n}{n+1}\]
B) \[\frac{n}{n+1}\]
C) \[\frac{2n}{n-1}\]
D) \[\frac{2n}{n+1}\]
Correct Answer: D
Solution :
| [d] Let the statement P(n) be defined as |
| \[P(n):1+\frac{1}{1+2}+\frac{1}{1+2+3}+...\] |
| \[+\frac{1}{1+2+3+...+n}=\frac{2n}{n+1}\] |
| 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 \[n=k+1,\] |
| \[\left( 1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{2}{k(k+1)} \right)+\frac{2}{(k+1)(k+2)}\] |
| \[=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}\] [Using equation (i)] |
| \[=\frac{2k(k+2)+2}{(k+1)(k+2)}=\frac{2\left[ {{k}^{2}}+2k+1 \right]}{(k+1)(k+2)}\] |
| [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. |
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