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

$$1+2+3+...+n<\dfrac{1}{8}(2n+1)^2$$

Asked by Pragya Singh | 1 year ago |  66

##### Solution :-

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

$$P(n): 1+2+3+...+n<\dfrac{1}{8}(2n+1)^2$$

For n=1,

$$1<\dfrac{1}{8}(2.1+1)^2=\dfrac{9}{8}$$

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

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

$$1+2+3+...+k<\dfrac{1}{8}(2k+1)^2$$ ..........(i)

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

Consider

$$1+2+3+...+k<\dfrac{1}{8}(2k+1)^2+(k+1)$$ [Using (i)]

$$<\dfrac{1}{8}\{(2k+1)^2+8(k+1)\}$$

$$<\dfrac{1}{8}\{4k^2+4k+1+8k+8\}$$

$$<\dfrac{1}{8}\{4k^2+12k+9\}$$

$$<\dfrac{1}{8}(2k+3)^2$$

$$<\dfrac{1}{8}\{2(k+1)+1\}^2$$
$$1+2+3+...+k+(k+1)$$

$$<\dfrac{1}{8}(2k+1)^2+(k+1)$$

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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