# Hard mathematical induction question

• Mathematica
Hi everyone, I have been trying to do this induction question... but can't seem to understand how to do it.

Prove by induction that, for all integers n is greater than or equal to 1,

(n+1)(n+2)....(2n-1)2n = 2^n[1x3x...x(2n-1)]

This is what someone else has given as the answer:

For n = 1
LHS = (1 + 1) = 2
RHS = 21 = 2
LHS=RHS hence statement is true for n = 1
Assume the statement is true for n = k
(k + 1)(k + 2).....(2k - 1)2k = 2k[1 x 3 x ......... x (2k - 1)]
Required to prove statement is true for n = k + 1
(k + 2)(k + 3).....(2k + 1)(2k + 2) = 2k + 1[1 x 3 x ......... x (2k + 1)]
LHS = (k + 2)(k + 3).....(2k + 1)(2k + 2)
= 2(k + 1)(k + 2)(k + 3).....2k(2k + 1)
= 2(2k + 1)2k[1 x 3 x ......... x (2k - 1)] by assumption
= 2k + 1[1 x 3 x ......... x (2k - 1) x (2k + 1)]
= RHS
If the statement is true for n = k, it is true for n = k + 1
Since the statement is true for n = 1, it is true for all positive integers n by induction.

I think I understand everything up until he actually makes the assumption. I don't understand the assumption and what he is actually subbing into the assumption.

Simon Bridge
Homework Helper
(k+1)(k+2)(k+3)...(2k-1)2k = 2k[1x3x5...x(2k-1)] = P(k)
... assumed true. (note: you left off the power?)

(k+2)(k+3)(k+4)...(2k+1)2(k+1) = 2(k+1)[1x3x5x...x(2k+1)] = P(k+1)
... needs to be proved. Kinda suggestive isn't it?

Consider:
Is the LHS of P(k+1) the same as the LHS of P(k)/(k+1) ?

The form of the LHS of P(n) is
f1f2f3...fm=Pm(k)

where
fi = k+i

but for i=m, fm = k+m = 2k(2k-1)
so what is fm-1 ?

Sorry, didn't realize the powers weren't copied properly.

Thanks for your reply, yea I get it now - I was looking at it from the wrong perspective.

Simon Bridge