Question

Prove by the principle of mathematical induction that 

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

Answer 1

Let T(n) be the statement 

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

Basic Step: 

For n = 1, we have \(\frac{(2n+1)^2}{8}\) = \(\frac{(2\times 1+1)^2}{8} = \frac{9}{8}\)  > 1 

⇒ T(1) is true.

Induction Step: 

Let T(k) be true. 

Then, 1 + 2 + ... + k <\(\frac{(2n+1)^2}{8}\)

Now we need to prove T(k + 1) to be true. To obtain T(k + 1) add the (k + 1)^th term = (k + 1) to both the sides of T(k). Then,

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

⇒ 1 + 2 + 3 + ... + k + (k + 1) < \(\frac{4k^2+12k+9}{8} =\frac{(2k+3)^2}{8}\) 

⇒ 1 + 2 + 3 + ... + k + (k + 1) < \(\frac{[2(k+1)+1]^2}{8}\) 

Hence T (k + 1) is true, whenever T(k) is true 

⇒ T(n) is true for all n∈N.