Use mathematical induction to prove that [a+(n-1)d] = (n/2)(a+an)?

Question

*note--- an= a sub n

I have figured part of this out, but I don't know what to do when I'm supposed to prove it with n+1 replacing n.

the actual thing that I am trying to prove is that
a + (a+d) + (a+2d) + (a+3d) +...+[a+(n-1)d] = (n/2)(a + a sub n)

Answer 1

Ah...

so, assuming sum(a_1 to a_n) = n(a+a_n)/2 for n, show that it's also true for n+1.

sum(a_1 to a_(n+1))

= sum(a_1 to a_n) + a_(n+1)

= n(a+a_n)/2 + (a+nd)

= [n(a + a + (n-1)d) + (2a + 2nd)] / 2

= [n(2a + nd) + (2a + nd)] / 2
(rearranging to get the form 2a + nd, which is a + a + nd = a + a_(n+1))

= (n+1)(a+a_(n+1)) / 2

QED. A fine case where the proof by induction is NOT the way to go. It's way easier to just show that the average of the numbers is the midpoint of the first and last number.

Answer 2

I think you've left some things out. [a + (n-1)d] is the formula for the nth term of an arithmetic series. (n/2)(a + a sub n) is the formula for the SUM of the first n terms of an arithmetic series. Those 2 will be equal ONLY when n=1, that is, when the series only has 1 term.

Answer 3

2n <= 2^n 2 * a million <=2^a million 2=2 2*2 <=2^2 4 = 4 2*3 <= 2^3 6 < 8 2*4 <= 2^4 8 < sixteen 2*5 <= 2^5 10 < 32 the the ideal option facet is growing to be speedier than the left, so it would desire to be actual. there is no data that the solar will upward push the next day. yet, it has risen daily so some distance. by potential of Induction, it quite is a stable wager that it's going to upward push the next day Do the 2d the comparable way.