Dont know why im having so much trouble with this i think im funking up the algebra. It doesn't actually say but i assume a and d are fixed and n is the variable to induct on.
a + (a + d) + (a + 2d) + ... + [a + (n-1)d] = n/2[2a + (n - 1)d]
Dont know why im having so much trouble with this i think im funking up the algebra. It doesn't actually say but i assume a and d are fixed and n is the variable to induct on.
a + (a + d) + (a + 2d) + ... + [a + (n-1)d] = n/2[2a + (n - 1)d]
You don't need to use induction.
Call this sum $\displaystyle S$.
Then
$\displaystyle S = a + (a + d) + (a + 2d) + \dots + [a + (n - 3)d] + [a + (n - 2)d] + [a + (n - 1)d]$
or
$\displaystyle S = [a + (n - 1)d] + [a + (n - 2)d] + [a + (n - 3)d] + \dots + (a + 2d) + (a + d) + a$.
Notice how I've written these terms in reverse on top of each other.
If I add up the the first term with the last, the second term with the second last, the third with the third last, etc...
$\displaystyle 2S = [2a + (n - 1)d] + [2a + (n - 1)d] + [2a + (n - 1)d] + \dots + [2a + (n - 1)d]$.
How many of these $\displaystyle 2a + (n - 1)d$ do we have?
We have as many of these as there are terms. I.e. we have $\displaystyle n$ of them.
So $\displaystyle 2S = n[2a + (n - 1)d]$.
Solving for $\displaystyle S$ we have
$\displaystyle S = \frac{n}{2}[2a + (n - 1)d]$.
I always used to write stuff out and people would just dump solutions down so i stopped lol heres what i got.
When n = 1, the base case is true a = 1/2(2a+0)=a so we get
a = (a+d) + ... + (a + (n-1)d) + (a + nd) = n/2[2a + (n-1)d] + (a + nd)
we need to get to (n+1)/2[2a + nd], so we get to cancel everything before the a + nd off and we get n/2[2a + (n-1)d] + (a+nd) = n/2[2a + (n-1)d] + (a + nd) after that i got it all over a common denominator and when i multiplied everything out i got to 2(2an) + d^2n - dn + 2d all over 2 and that is not where im supposed to get to (i multiplied everything out since i suck at factoring huge stuff like that) and obviously all the d's should cancel out. Did i flip a sign somewhere? I tried it a few times and got the same answer. Thanks for your help
wow i would have never thought of doing it that way is it just experience that teaches you how to tackle some of this stuff or am i just dumb? id like to think im pretty smart but being a math major is a good way to make anyone feel like an idiot from time to time. I know you dont have to but it is possible to do this problem with induction isnt it? ( i know it may be more difficult)
To show that
$\displaystyle a + (a + d) + (a + 2d) + \dots + [a + (n - 1)d] = \frac{n}{2}[2a + (n - 1)d]$...
Base step: $\displaystyle n = 1$
$\displaystyle LHS = a$
$\displaystyle RHS = \frac{1}{2}[2a + (1 - 1)d]$
$\displaystyle = \frac{1}{2}(2a)$
$\displaystyle = a$
$\displaystyle = LHS$.
Inductive step:
Assume the equation is true for $\displaystyle n = k$.
So we assume
$\displaystyle a + (a + d) + (a + 2d) + \dots + [a + (k - 1)d] = \frac{k}{2}[2a + (k - 1)d]$.
Now we show the equation is true for $\displaystyle n = k + 1$.
I.e. we show that
$\displaystyle a + (a + d) + (a + 2d) + \dots + [a + (k - 1)d] + (a + kd) = \frac{k + 1}{2}[2a + kd]$.
$\displaystyle LHS = a + (a + d) + (a + 2d) + \dots + [a + (k - 1)d] + (a + kd)$
$\displaystyle = \frac{k}{2}[2a + (k - 1)d] + (a + kd)$
$\displaystyle = \frac{2ak}{2} + \frac{k(k - 1)d}{2} + \frac{2a}{2} + \frac{2kd}{2}$
$\displaystyle = \frac{1}{2}[2ak + k(k - 1)d + 2a + 2kd]$
$\displaystyle = \frac{1}{2}[2ak + 2a + (k^2 - k + 2k)d]$
$\displaystyle = \frac{1}{2}[2a(k + 1) + (k^2 + k)d]$
$\displaystyle = \frac{1}{2}[2a(k + 1) + k(k + 1)d]$
$\displaystyle = \frac{k + 1}{2}[2a + kd]$
$\displaystyle = RHS$.
Q.E.D.