Prove that 1^2−2^2+3^2−···+(−1)^(n−1)n^2=(−1)^(n−1) n(n + 1)/2 whenever n is a positive integer

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:

1. BASIS STEP: We verify that P (1) is true.
2. 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 a^m-n=a^m/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.

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