C2 Sequences and series.

Hey.
Please can someone explain how i can use Sn = n/2( 2a + (n-1)d) to find the sum of consecutive numbers, consecutive even/odd numbers, or point me to where i can find out.
I don`t quite understand how to manipulate the equation.
Thanks in advanced.

Reply 1

Rager6amer

Original post by SamuelN98

how many numbers are we talking?

Reply 2

Kevin De Bruyne

Original post by SamuelN98

In each case, work out what n is, work out what a is and work out what d is.

Eg keeping it simple, how do you find the sum of 1,2,3? How about 1,3,5? 2,4,6?

Reply 3

SamuelN98

Original post by SeanFM

In each case, work out what n is, work out what a is and work out what d is.

Eg keeping it simple, how do you find the sum of 1,2,3? How about 1,3,5? 2,4,6?

Thanks. i`m still finding it difficult still to apply it to summation of larger series

S3=(1/2)(3)(2+(3-1)1)

S3=(1/2)(3)(2+(3-1)2)

S3=(1/2)(3)(2(2)+(3-1)2)

Do i just derive the 3 formula above, then sub in a different N value?

Reply 4

Kevin De Bruyne

Original post by SamuelN98

Yes, those 3 are correct. An example of something that you're struggling and how you've tried to tackle it would help :h:

I'm not sure what you mean by that last bit.

Reply 5

SamuelN98

Original post by SeanFM

Yes, those 3 are correct. An example of something that you're struggling and how you've tried to tackle it would help :h:

I'm not sure what you mean by that last bit.

Sorry, i just meant if a question asked for the sum of the the even numbers from 0 to 100 etc, what value of N would i use.

Ill try and find an example.
Thanks.

Reply 6

Kevin De Bruyne

Original post by SamuelN98

Sorry, i just meant if a question asked for the sum of the the even numbers from 0 to 100 etc, what value of N would i use.

Ill try and find an example.
Thanks.

Oh yes, I know what you mean - for example - the sum of all even numbers from 2 to 100 inclusive.

Right, a and d are straightforward - they are both 2.

But n is slightly less easier. What I would suggest is thinking about it / using your fingers like this:

'2 is the first number, 4 is the second, 6 is the third, 8 is the fourth, 10 is the fifth... 12 is the sixth, 14 is the seventh, 16 is the eighth, 18 is the ninth and 20 is the 10th. So the one that ends in '10' are multiples of 5, is what we learned from this set of numbers. So using that, 30 is the 15th, 40 is the 20th, 50 is the 25th.. double that and you get that 100 is the 50th term. So n is 50 in this case.

Reply 7

Zacken

As a more formulaic alternative:

a = 2 + (n-1)(2) = 2n

is the sequence that generates even numbers.

So the 100 th term is when

100 = 2n \iff n = 50

Reply 8

SamuelN98

Original post by SeanFM

Cheers, very helpful.
Would i use a similar idea for Odd numbers?

Reply 9

Kevin De Bruyne

Original post by SamuelN98

Cheers, very helpful.
Would i use a similar idea for Odd numbers?

Challenge for you then, what is the sum of the odd numbers from 2 to 50?

Reply 10

SamuelN98

Original post by SeanFM

Challenge for you then, what is the sum of the odd numbers from 2 to 50?

sn=(n/2)(2a+(n-1)d)

S23= (23/2)((2(3)+(22)(2))=690 ???

Reply 11

Kevin De Bruyne

Original post by SamuelN98

sn=(n/2)(2a+(n-1)d)

S23= (23/2)((2(3)+(22)(2))=690 ???

The key thing is getting a, n and d right :tongue:

I am sure you can put it in the formula correctly. (and of course, finding n is the hardest bit).

Again, 3 is the first term, 5 is the second, 7 is the third, 9 is the fourth, 11 is the fifth, 13 is the sixth... 23 is the eleventh term, 43 is the twenty first term, 45 is twenty two, 47 is twenty three and 49 is twenty four. So n = 24 in this case.