Answer:

Solution:

From the statement formula

When n = 1,

LHS =1 x 2 = 2

RHS = 1(1+1)(4x1−1)3 = 63 = 2

Hence it is proved that P (1) is true for the equation.

Now we assume that P (k) is true or 1 x 2 + 3 x 4 + 5 x 6 + …. + (2k - 1) x 2k = k(k+1)(4k−1)3.

For P(k + 1)

LHS = 1 x 2 + 3 x 4 + 5 x 6 + …. + (2k - 1) x 2k + (2(k + 1) - 1) x 2(k + 1)

= k(k+1)(4k−1)3 + (2(k + 1) - 1) x 2(k + 1)

= (k+1)3(4k2 - k + 12 k + 6)

= (k+1)(4k2+8k+3k+6)3

= (k+1)(k+2)(4k+3)3

= (k+1)((k+1)+1)(4(k+1)−1)3 = RHS for P (k+1)

Now it is proved that P (k + 1) is also true for the equation.

So the given statement is true for all positive integers.