Where you stuck? Please show your effort so that we can help you in a better way.
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]
Call this sum .
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...
How many of these do we have?
We have as many of these as there are terms. I.e. we have of them.
Solving for we have
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)