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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.