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

$$1.3+3.5+5.7+....+(2n-1)(2n+1)$$

$$=\dfrac{n(4n^2+6n-1)}{3}$$

Asked by Pragya Singh | 1 year ago |  47

#### 1 Answer

##### Solution :-

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

$$1.3+3.5+5.7+....+(2n-1)(2n+1)$$

$$=\dfrac{n(4n^2+6n-1)}{3}$$

For n=1,
L.H.S. = 1.3 = 3

$$R.H.S =\dfrac{1(4.1^2+6.1-1)}{3}$$

$$=\dfrac{n(4+6-1)}{3}$$

$$=\dfrac{9}{3}=3$$

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

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

$$1.3+3.5+5.7+....+(2k-1)(2k+1)$$

$$=\dfrac{k(4k^2+6k-1)}{3}$$.........(i)

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

Consider

$$1.3+3.5+5.7+....+(2k-1)(2k+1)$$

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

$$\dfrac{k(4k^2+6k-1)}{3}+(2k+2-1)(2k+2+1)$$

$$\dfrac{k(4k^2+6k-1)}{3}+(2k+1)(2k+3)$$

$$\dfrac{k(4k^2+6k-1)}{3}+(4k^2+8k+3)$$

$$\dfrac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}$$

$$\dfrac{4k^2+18k^2+23k+9}{3}$$

$$\dfrac{4k^2+14k^2+9k+4k^2+14k+9}{3}$$

$$\dfrac{k(4k^2+14k^2+9)+1(4k^2+14k+9)}{3}$$

$$\dfrac{(k+1)\{4(k^2+2k+1)+6(k+1)-1\}}{3}$$

$$\dfrac{(k+1)\{4(k+1)^2+6(k+1)-1\}}{3}$$

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 Abhisek | 1 year ago

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