# Math Help - Simple Induction Problem

1. ## Simple Induction Problem

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]

2. Where you stuck? Please show your effort so that we can help you in a better way.

3. Originally Posted by ChrisBickle
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 $S$.

Then

$S = a + (a + d) + (a + 2d) + \dots + [a + (n - 3)d] + [a + (n - 2)d] + [a + (n - 1)d]$

or

$S = [a + (n - 1)d] + [a + (n - 2)d] + [a + (n - 3)d] + \dots + (a + 2d) + (a + d) + a$.

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...

$2S = [2a + (n - 1)d] + [2a + (n - 1)d] + [2a + (n - 1)d] + \dots + [2a + (n - 1)d]$.

How many of these $2a + (n - 1)d$ do we have?
We have as many of these as there are terms. I.e. we have $n$ of them.

So $2S = n[2a + (n - 1)d]$.

Solving for $S$ we have

$S = \frac{n}{2}[2a + (n - 1)d]$.

4. 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

5. 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)

6. Originally Posted by ChrisBickle
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)
Most text books provide the simple proof to the sum of an arithmetic sequence that I show you. And yes, experience helps too.

Induction is possible, but quite a lot more work.

7. To show that

$a + (a + d) + (a + 2d) + \dots + [a + (n - 1)d] = \frac{n}{2}[2a + (n - 1)d]$...

Base step: $n = 1$

$LHS = a$

$RHS = \frac{1}{2}[2a + (1 - 1)d]$

$= \frac{1}{2}(2a)$

$= a$

$= LHS$.

Inductive step:

Assume the equation is true for $n = k$.

So we assume

$a + (a + d) + (a + 2d) + \dots + [a + (k - 1)d] = \frac{k}{2}[2a + (k - 1)d]$.

Now we show the equation is true for $n = k + 1$.

I.e. we show that

$a + (a + d) + (a + 2d) + \dots + [a + (k - 1)d] + (a + kd) = \frac{k + 1}{2}[2a + kd]$.

$LHS = a + (a + d) + (a + 2d) + \dots + [a + (k - 1)d] + (a + kd)$

$= \frac{k}{2}[2a + (k - 1)d] + (a + kd)$

$= \frac{2ak}{2} + \frac{k(k - 1)d}{2} + \frac{2a}{2} + \frac{2kd}{2}$

$= \frac{1}{2}[2ak + k(k - 1)d + 2a + 2kd]$

$= \frac{1}{2}[2ak + 2a + (k^2 - k + 2k)d]$

$= \frac{1}{2}[2a(k + 1) + (k^2 + k)d]$

$= \frac{1}{2}[2a(k + 1) + k(k + 1)d]$

$= \frac{k + 1}{2}[2a + kd]$

$= RHS$.

Q.E.D.