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 agoGiven an example of a statement P (n) such that it is true for all n ϵ N.

a + (a + d) + (a + 2d) + … + (a + (n-1)d) = \( \dfrac{n}{2}\) [2a + (n-1)d]

7^{2n} + 2^{3n – 3}. 3n – 1 is divisible by 25 for all n ϵ N

n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.

(ab)^{ n} = a^{n} b^{n} for all n ϵ N