mathematical induction Q help (1 Viewer)

[addoil]

prove:
a + (a+d) + (a+2d)...+ [a+d(n-1)] = (n/2)[2a+d(n-1)]
(ie. prove the arithmetic sum formula!)

Here's what I have:
-----------------------

When [n=1] LHS = a, RHS = a
Therefore the statement is true for n=1

Assume the statement is true for n=k;
ie. Sum(k) = (k/2) [2a+d(k-1)]

Therefore prove that the statement is true for n=k+1
ie. Sum(k+1) = [(k+1)/2] (2a+dk)
LHS
= S(k+1)
= S(k) + T(k+1)
= (k/2) [2a+d(k-1)] + [a+dk]
= (k/2) [2a+dk-d] + (1/2) [2a+2dk]
= [(k+1)/2] [2a+dk-d+2a+2dk]
= [(k+1)/2] [4a+3dk-d]

I got the first part which is the (k+1)/2 part... but how do i prove the 2a+dk part?!?!?

 

[Aesytic]

= (k/2) [2a+dk-d] + (1/2) [2a+2dk]
= [(k+1)/2] [2a+dk-d+2a+2dk]
that line was incorrect, you didn't actually factorise that properly
(k/2)[2a+dk-d] + (1/2)[2a+2dk] = (1/2)[k(2a+dk-d)] + (1/2)[2a+2dk]
= (1/2)[k(2a+dk-d) + 2a + 2dk]
= (1/2)[2ak + dk^2 - dk + 2a + 2dk]
= (1/2)[2ak + dk^2 + 2a + dk]
= (1/2)[2a(k+1) + dk(k+1)]
= (1/2)[(k+1)(2a + dk)]
= ((k+1)/2)[2a+dk]
=RHS

 

[umm what]

+1

= (k/2) [2a+dk-d] + (1/2) [2a+2dk]
= [(k+1)/2] [2a+dk-d+2a+2dk]
that line was incorrect, you didn't actually factorise that properly
(k/2)[2a+dk-d] + (1/2)[2a+2dk] = (1/2)[k(2a+dk-d)] + (1/2)[2a+2dk]
= (1/2)[k(2a+dk-d) + 2a + 2dk]
= (1/2)[2ak + dk^2 - dk + 2a + 2dk]
= (1/2)[2ak + dk^2 + 2a + dk]
= (1/2)[2a(k+1) + dk(k+1)]
= (1/2)[(k+1)(2a + dk)]
= ((k+1)/2)[2a+dk]
=RHS