Answer:
We will use mathematical induction to prove the formula for the sum of arithmetic sequences:
For n=1, we have:
∑j=1^1(a + (j-1)d) = a
On the other hand, we have:
n/2(2a + (n-1)d) = 1/2(2a) = a
Thus, the formula holds for n=1.
Assuming the formula holds for n=k, we will prove that it holds for n=k+1.
We have:
∑j=1^(k+1)(a + (j-1)d) = (a + kd) + ∑j=1^k(a + (j-1)d)
Using the formula for n=k, we can write:
∑j=1^k(a + (j-1)d) = k/2(2a + (k-1)d)
Substituting this back into the first equation, we have:
∑j=1^(k+1)(a + (j-1)d) = (a + kd) + k/2(2a + (k-1)d)
Simplifying the right-hand side, we get:
∑j=1^(k+1)(a + (j-1)d) = 1/2(2a + (2k+1)d)
But (k+1)/2(2a + kd + d) = 1/2(2a + (2k+1)d), so the formula holds for n=k+1.
Therefore, by mathematical induction, the formula for the sum of arithmetic sequences is proved.