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

\( 1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+.......\)

\( +\dfrac{1}{1+2+3+..n}=\dfrac{2n}{(n+1)}\)

Asked by Pragya Singh | 2 years ago |  88

1 Answer

Solution :-

Let us denote the given equality by P(n), i.e.,

\( 1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+.......\)

\( +\dfrac{1}{1+2+3+..n}=\dfrac{2n}{(n+1)}\)

For n=1,

L.H.S.= \( \dfrac{1}{1}=1\)

R.H.S.= \( \dfrac{2.1}{1+1}=1\)

Therefore, P(n) is true for n=1.

Let us assume that P(k) is true for some positive integer k , i.e.,
\( 1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+.......\)

\( +\dfrac{1}{1+2+3+..k}=\dfrac{2k}{(k+1)}\).........(i)

Now, we have to prove that P( k+1) is also true.

Consider

\( 1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+.......\)

\( +\dfrac{1}{1+2+3+......+k}\)

\( =\dfrac{1}{1+2+3+......+k+(k+1)}\)

\( \dfrac{2k}{k+1}+\dfrac{1}{1+2+3}+...+k(k+1)\) .........[using (i)]\(\)

\( \dfrac{2k}{k+1}+\dfrac{1}{(\dfrac{(k+1)(k+1+1)}{2})}\)

\( \dfrac{2k}{k+1}+\dfrac{2}{(k+1)(k+2)}\)

\( \dfrac{2k}{k+1}(k+\dfrac{1}{k+2})\)

\( \dfrac{2k}{k+1}[\dfrac{(k+1)^2}{k+2}]\)

\( \dfrac{2(k+1)}{k+2}\)

Therefore, P(k+1) holds whenever P(k) holds.

Hence, the given equality is true for all natural numbers i.e., N by the principle of mathematical induction.

Answered by Pragya Singh | 2 years ago

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