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URGENT HELP NEEDED!!! Proof by induction

Hi guys. Could anyone please help me to solve this question?? It's sort of urgent :smile:

Prove by Induction:

Sum of terms of an arithmetic sequence
Sn = n/2(2a + (n-1)d)

Thank you muchos! :wink:
Original post by medicng
Hi guys. Could anyone please help me to solve this question?? It's sort of urgent :smile:

Prove by Induction:

Sum of terms of an arithmetic sequence
Sn = n/2(2a + (n-1)d)

Thank you muchos! :wink:


Do you know how to start? How would you usually do an induction?
Reply 2
Original post by hassi94
Do you know how to start? How would you usually do an induction?


I usually start with n= 1 , then assume for n = k , and finish with n = k + 1 but I got stuck at the start with this one.
Original post by medicng
I usually start with n= 1 , then assume for n = k , and finish with n = k + 1 but I got stuck at the start with this one.


Well okay, when n = 1, S1=12(2a1+0d)=12(2a1)=a1S_1 = \frac{1}{2}(2a_1 + 0d) = \frac{1}{2}(2a_1) = a_1 which is obviously true as the sum of the first 1 terms is just equal to the first term.

Now, you assume n=k is correct, so assume:

Sk=k2(2a1+(k1)d)S_k = \frac{k}{2}(2a_1 + (k-1)d)

is true.

Then you get Sk+1=S_{k+1} =
Fill in the right hand side, bearing in mind you know what the sum of the first k terms is equal to (you've assumed it's true).

Also remember the formula an=a1+(n1)da_n = a_1 + (n-1)d.
Reply 4
Original post by hassi94
Well okay, when n = 1, S1=12(2a1+0d)=12(2a1)=a1S_1 = \frac{1}{2}(2a_1 + 0d) = \frac{1}{2}(2a_1) = a_1 which is obviously true as the sum of the first 1 terms is just equal to the first term.

Now, you assume n=k is correct, so assume:

Sk=k2(2a1+(k1)d)S_k = \frac{k}{2}(2a_1 + (k-1)d)

is true.

Then you get Sk+1=S_{k+1} =
Fill in the right hand side, bearing in mind you know what the sum of the first k terms is equal to (you've assumed it's true).

Also remember the formula an=a1+(n1)da_n = a_1 + (n-1)d.


Thanks :smile: I'll try it and hopefully won't mess it up

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