If k = 1, then k+1 = 2:
LHS = (1 – 1/4) = 3/4
RHS = (1+2)/2(1+1) = 3/4

If k = 2, then k+1 = 3:
LHS = (1 – 1/4)(1 – 1/9) = 3/4 * 8/9 = 2/3
RHS = (2+2)/2(2+1) = 4/6 = 2/3

If k = 3, then k+1 = 4:
LHS = (3/4)(8/9)(15/16) = 5/8
RHS = (3+2)/2(3+1) = 5/8

If k = 4, then k+1 = 5:
LHS = (3/4)(8/9)(15/16)(24/25) = 3/5
RHS = (4+2)/2(4+1) = 6/10 = 3/5

Pattern is 3/4, 4/6, 5/8, 6/10, etc., with the numerators going up 1 while denominators go up by 2.

So the equality seems to be true so far. It has been established for k = 1, 2, 3, 4.

The induction step requires you to prove that if the equation is known to be true for k = n (some arbitrary whole number), then it must also be true for k = n+1.

So what you do is write out the equation using n+1 everywhere k appears. Then try to prove that but it’s okay to use the immediately previous result.

We seek to prove:

(1 – 1/2²)…(1 – 1/(n+1)²)(1 – 1/(n+1+1)²)

= (n+1+2)/2(n+1+1)

= (n+3)/2(n+2)

We can use the fact that the equation is true for k = n. So left hand simplifies to a simple expression already proved, multiplied by just one new term:

LHS = (n+2)/(2(n+1)) * (1 – 1/(n+2)²)

= (n+2)/(2(n+1)) * ((n+2)² – 1)/(n+2)²

Cancel (n+2)/(n+2) in the product.

= [(n+2)² – 1] / [2(n+1)(n+2)]

Use difference of squares.

= (n+2 + 1)(n+2 – 1) / 2(n+1)(n+2)

= (n+3)(n+1) / 2(n+1)(n+2)

Cancel (n+1)/(n+1).

= (n+3) / 2(n+2)

QED

You’ve shown it’s true for k = 1; and you’ve now seen that if it’s true for k = n, then it is true for k = n+1. Therefore it is true for any whole number k.