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

$$(1+ \dfrac{3}{1})(1+\dfrac{5}{4})(1+\dfrac{7}{9})....$$

$$(1+\dfrac{(2n+1)}{n^2})=(n+1)^2$$

Asked by Abhisek | 1 year ago |  64

#### 1 Answer

##### Solution :-

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

$$(1+ \dfrac{3}{1})(1+\dfrac{5}{4})(1+\dfrac{7}{9})....$$

$$(1+\dfrac{(2n+1)}{n^2})=(n+1)^2$$

For n=1,

L.H.S.= $$(1+ \dfrac{3}{1})=4$$

R.H.S.= (1+1)2 = 22 = 4

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

Let us assume that P(k) is true for some positive integer k , i.e.,

$$(1+ \dfrac{3}{1})(1+\dfrac{5}{4})(1+\dfrac{7}{9})....$$

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

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

Consider

$$[ (1+ \dfrac{3}{1})(1+\dfrac{5}{4})(1+\dfrac{7}{9})....$$

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

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

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

$$(k+1)^2+2(k+1)+1$$

$$\{(k+1)+1\}^2$$

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 | 1 year ago

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