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  • Prove the following by using the principle of mathematical induction for all n ∈ N: (3) 1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...n)=2n/(n+1)

Prove the following by using the principle of mathematical induction for all n\in N:

Q : 3        1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+n)}=\frac{2n}{(n+1)}

Answers (1)

best_answer

Let the given statement be p(n) i.e.
p(n):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+n)}=\frac{2n}{(n+1)}
For n = 1  we have
p(1):1=\left (\frac{2(1)}{1+1} \right )= \left ( \frac{2}{2} \right )=1    ,   which is true

For  n = k  we have
p(k):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k)}=\frac{2k}{(k+1)} \ \ \ \ -(i)   ,        Let's assume that this statement is true

Now,
For  n = k + 1  we have
p(k+1):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k+1)}                                                                                                                                                                                        =\left ( 1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k)} \right )+\frac{1}{(1+2+3+...+k+k+1)}
                                                                 
                                                                                  =\frac{2k}{k+1}+\frac{1}{(1+2+3+...+k+(k+1))} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
                                                                                  =\frac{2k}{k+1}+\frac{1}{\frac{(k+1)(k+1+1)}{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because1+2+....+n = \frac{n(n+1)}{2} )
                                                                                  =\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}
                                                                                  =\frac{2}{k+1}\left (k+\frac{1}{k+2} \right )
                                                                                  =\frac{2}{k+1}\left ( \frac{k^2+2k+1}{k+2} \right )
                                                                                  =\frac{2}{k+1}.\frac{(k+1)^2}{k+2}
                                                                                  =\frac{2(k+1)}{k+2}
Thus,  p(k+1)  is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n)  is true for all natural numbers n

Posted by

Gautam harsolia

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