Answer:
Explanation:
To prove the formula for the sum of an arithmetic series, we need to show that it holds for the base case P(1) and then demonstrate that if it holds for P(k), it also holds for P(k+1).
Step 1: Prove P(1)
For the base case P(1), we have:
Σ(α + (j − 1)d) from j=1 to 1 = (1/2) * (2α + (1 - 1)d)
On the left side, the sum only includes one term (when j=1), so it simplifies to:
(α + (1 − 1)d)
On the right side, the formula becomes:
(1/2) * (2α + (1 - 1)d)
Now, let's evaluate both sides:
Left side: (α + (1 − 1)d) = α
Right side: (1/2) * (2α + (1 - 1)d) = (1/2) * (2α) = α
Since both sides are equal (both are α), the formula holds for P(1).
Step 2: Prove P(k+1) assuming P(k) is true
Now, let's assume that the formula holds for P(k), where k is a positive integer. That is:
Σ(α + (j − 1)d) from j=1 to k = (k/2) * (2α + (k - 1)d)
We want to prove that the formula also holds for P(k+1). That is:
Σ(α + (j − 1)d) from j=1 to k+1 = ((k+1)/2) * (2α + (k + 1 - 1)d)
To do this, let's add the next term (the (k+1)th term) to the sum on the left side:
Σ(α + (j − 1)d) from j=1 to k+1 = Σ(α + (j − 1)d) from j=1 to k + (α + (k+1 - 1)d)
Using the assumption P(k):
Σ(α + (j − 1)d) from j=1 to k + (α + (k+1 - 1)d) = (k/2) * (2α + (k - 1)d) + (α + kd)
Now, let's simplify the expression:
(k/2) * (2α + (k - 1)d) + (α + kd) = (k/2) * (2α + kd - d) + (α + kd)
= (k/2) * (2α + kd) - (k/2) * d + (α + kd)
= kα + (k^2/2)d - (k/2) d + α + kd
= kα + (k^2/2)d + α + kd - (k/2) d
= kα + α + kd + (k^2/2)d - (k/2) d
= (kα + α) + (kd - (k/2)d) + (k^2/2)d
= α(k + 1) + d(k - (k/2) + (k^2/2))
= α(k + 1) + d(k - k/2 + k^2/2)
= α(k + 1) + d(k/2 + k^2/2)
= α(k + 1) + d(k/2)(1 + k)
= α(k + 1) + d(k/2)(k + 1)
= (α + (k/2)d)(k + 1)
Now, notice that (α + (k/2)d) is the (k+1)th term of the arithmetic series (since a = α, d = d). Therefore, the sum on the left side is:
Σ(α + (j − 1)d) from j=1 to k+1 = (α + (k/2)d)(k + 1)
But, we know from the formula for the sum of an arithmetic series:
Σ(α + (j − 1)d) from j=1 to k+1 = ((k+1)/2) * (2α + (k + 1 - 1)d)
Comparing both sides, we have:
(α + (k/2)d)(k + 1) = ((k+1)/2) * (2α + (k + 1 - 1)d)
Since both sides are equal, we have shown that if the formula holds for P(k), it also holds for P(k+1).
Conclusion:
We have proven the formula for the sum of an arithmetic series, both for the base case P(1) and for the inductive step P(k+1) assuming P(k) is true. Therefore, the formula is true for all positive integers n.