We should always start with the definitions when solving exercises. So, let’s see the principle of mathematical induction.
PRINCIPLE OF MATHEMATICAL INDUCTION To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps:
- BASIS STEP: We verify that P (1) is true.
- INDUCTIVE STEP: We show that the conditional statement P (k) → P (k + 1) is true for all positive integers k.
The definition above is from the textbook Discrete Mathematics and its Applications by Rosen.
Now, let’s solve the exercise.
Prove that 12−22+32+···+(−1)n-1n2=(−1)n-1n(n + 1)/2, whenever n is a positive integer
Basis step:
P(1): 12 = (−1)1-11(1 + 1)/2=-101(2)/2 = 1
Inductive step:
P(k): 12−22+32+···+(−1)k-1k2 = (−1)k-1k(k + 1)/2
P(k+1): 12−22+32+···+(−1)k-1k2 + (−1)(k+1)-1(k+1)2 = (−1)(k+1)-1(k+1)(k+1 + 1)/2
12−22+32+···+(−1)k-1k2 + (−1)k(k+1)2 = (−1)k (k+1)(k+2)/2
Substituting P(k) in P(k+1) we get,
(−1)k-1k(k + 1)/2 + (−1)k(k+1)2 = (−1)k (k+1)(k+2)/2
(−1)k-1k(k + 1)/2 + 2(−1)k(k+1)2/2 = (−1)k (k+1)(k+2)/2
Notice that am-n=am/an, therefore, (-1)k-1=(-1)k/(-1)1=-(-1)k
(-1)k(k+1)(-k+2(k+1))/2 = (−1)k (k+1)(k+2)/2
(-1)k(k+1)(-k+2k+2))/2 = (−1)k (k+1)(k+2)/2
(-1)k(k+1)(k+2))/2 = (−1)k (k+1)(k+2)/2 ⧠
Here is another example of previous knowledge you should have. If you don’t know the properties of exponentiation, you won’t be able to finish this proof.
Related pots:
- a) Find a formula for the sum of the first n even positive integers. Prove the formula that you conjectured in part (a)
- Prove that 1·1!+2·2!+···+n·n!=(n+1)!−1 whenever n is a positive integer.
- Prove that 1^2 +3^2 +5^2 +···+(2n+1)^2 =(n+1) (2n + 1)(2n + 3)/3 whenever n is a nonnegative integer
- Mathematical induction with examples