Question

In my past account, no one was able to solve this. I wonder if someone can solve this now.

Prove:

\[S_n = \frac n2 (2a + (n-1)d)\]

Answer 1

no substitution allowed (by that i mean brutely substituting every number to prove it's right) and no mathematical induction either

Answer 2

Sn being the sum of any arithmetic progression??

Answer 3

it is

Answer 4

a1+a2+a3+a4+a5+....+an=a1+(a1+d)+(a1+2d)+......+(a1+(n-1)d)
then, using conmutative property
Sn=a1+a2+a3+.....+an= (a1+a1+a1+a1+a1+a1+a1+a1+..+a1)+(d+2d+.....(n-1)d)
Sn=a1*n+(d+2d+....(n-1)d)
factorize by d
Sn=n*a1+d(1+2+3+....(n-1))
which is the sum of the first n-1 natural numbers (which is equal to (n-1)n/2 )
Sn=n*a1+d(n(n-1)/2)
\[S_n=n(a1+\frac{ d }{ 2 } (n-1))\] then just multiply and divide by 2 on the right side
\[S_n=\frac{ n }{ 2 } (2a_1+d(n-1))\]
works for you?

Answer 5

isn't this a variation of mathematical induction?

Answer 6

but not induction itself xD

Answer 7

lol. unfair :p well i'll see first if anyone can solve it without mathematical induction

Answer 8

Use what Gauss used to solve the 1+2+...+100 problem: Write the sum twice, once forward, once backward.

Answer 9

oh that thing?

Answer 10

To elaborate:
\[S=a+(a+d)+(a+2d)+\dots+(a+(n-1)d)\\
S=(a+(n-1)d)+(a+(n-2)d)+\dots+(a+d)+a\]
Add them, and notice that there are n terms on RHS. Each of these add up to \(2a+(n-1)d\):
\[2S=n(2a+(n-1)d)\]
and the result follows.

Answer 11

but...isn't this to infinity? how do you divide infinity by 2?

Answer 12

ah. that

Answer 13

hmm yes. you cross multiplied. then what?

Answer 14

you guys are genius. wish have more medal to give

Answer 15

Well, it's not obvious because I don't know how to align things. The first term of the RHS in the last sum is \(a+(a+(n-1)d)=2a+(n-1)d\), the next term is \((a+d)+(a+(n-2)d=2a+(n-1)d\), etc. There are \(n\) of these, in total.

Answer 16

Does this work for you?

Answer 17

s = a + a+d +.....a + (n - 2)d + a + (n -1)d
now write s in reverse order...
s = a + (n-1)d + a + (n-2)d
Now add them....term by term
s = a + a + d + a + (n-2)d + a + (n-1)d
s = a + (n-1)d + a + (n-2)d + a + d + a
-----------------------------------------------------------
2s = (2a + (n-1)d + (2a + (n-1)d + (2a + n-1)d......+ (2a + (n-1)d

Each term is the same and there are " n " of them so :
s = n x (2a + (n-1)d)

so just divide by 2 and we get :
s = n/2 x (2a (n-1)d)

Answer 18

oops forgot something....at the top, in reverse order
s = a + (n-1)d + a + (n-2)d + (a + d) + a

Answer 19

what does s mean in your solution texas?

Answer 20

is it also sum?

Answer 21

yes..it is sum

Answer 22

i must admit...these three are smarter than 3 green users =))) 3 green users were unable to solve it in the past

Answer 23

it can be confusing :)

Answer 24

That's what reading The Art and Craft of Problem Solving does to you.

Answer 25

so true :)

Answer 26

i would show you what they answered....but that would mean revealing the identity of my permanently banned account...so i cant...especially in the presence of a mod :S

Answer 27

Yes, I agree.....I wouldn't take any chances