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If you understand arithmetic progression that result should not need further explanation. The link provided gives a proof of the arithmetic series, you should read it.
In this sequence, Sn = 1 + 3 + 5 + ...+ 2n-1, the first item is 1, the last item is (2n-1), there are altogether n items in the sequence. If you are asking how to sum all the items in an arithmetic progression, I can give you an example, 1+ 2 + ... + 100, you can see that 1+100 = 101, 2 + 99 = 101, 3 + 98 = 101, ... 49+ 52 = 101, 50 + 51 = 101, so you just need to pair the first and last item, and how many pairs are there? 50 pairs, so the sum is (1+100) * 100/ 2; I cannot think any more to explain this to you but I really think that link is sufficient enough
It's actually defined as n(n+1)/2, because if the series has an odd number of elements, then the center element needs to be added in hence:
summish(3) = 1 + 1 + 2 + 2 + 3 + 3 / 2 = (3)(3+1) / 2 = 12 / 2 = 6
To find the sum of numbers in an airthmetic progression,you have a defined formula which can be proved. sum of progression = n/2*(1st term + last term); where n stands for total number of terms in that series.
Is there a need to calculate for Tn = 2n-1? As far as I can see the sequence resolves to Sn = n2 as all the other terms cancel out and (n-1)2 for t1 is zero. Wouldn't this give constant time solution? Or am I missing something?
formula of sum of AP series (which is 1,3,5,...,2*n-1)
n/2*(a+(n-1)*d)
given d=2 and a=1(as a first term and d is difference between two numbers in the series)
which makes above equation n*n
Summing the N series
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NO NEED to calculate every term, notice that Tn = 2n-1, thus Sn = 1 + 3 + 5 + ...+ 2n-1= (1+2n-1) * n / 2 = n*n.
For those wondering why,
Okay, I get why
T_n = 2n - 1, basic square expansion. But from that, how do you calculateΣ_n?have no idea what you are trying to do...the result is n*n, that's all
I'm trying to understand where does this
n*ncomes from / how did you get there , connecting the dots from the fact thatT_n = 2n - 1till your resultthat's what I don't get how do you reach to this
https://en.wikipedia.org/wiki/Arithmetic_progression
Thanks, I know what an arithmetic progression is but could you explain how do you get this
(1+2n-1) * n / 2If you understand arithmetic progression that result should not need further explanation. The link provided gives a proof of the arithmetic series, you should read it.
But! If than answer isn't sufficient for you... proof by induction.
very nice, i havent seen a proof by induction in ages!
Yeah, I like this proof better too... The other one is just overcomplicating things
In this sequence, Sn = 1 + 3 + 5 + ...+ 2n-1, the first item is 1, the last item is (2n-1), there are altogether n items in the sequence. If you are asking how to sum all the items in an arithmetic progression, I can give you an example, 1+ 2 + ... + 100, you can see that 1+100 = 101, 2 + 99 = 101, 3 + 98 = 101, ... 49+ 52 = 101, 50 + 51 = 101, so you just need to pair the first and last item, and how many pairs are there? 50 pairs, so the sum is (1+100) * 100/ 2; I cannot think any more to explain this to you but I really think that link is sufficient enough
Thanks, I had forgotten there's a rule that the sum of a progression is always n*(firstelement+lastelement)/2 . Now things are clear.
It's actually defined as n(n+1)/2, because if the series has an odd number of elements, then the center element needs to be added in hence: summish(3) = 1 + 1 + 2 + 2 + 3 + 3 / 2 = (3)(3+1) / 2 = 12 / 2 = 6
To find the sum of numbers in an airthmetic progression,you have a defined formula which can be proved. sum of progression = n/2*(1st term + last term); where n stands for total number of terms in that series.
Elementary mathematics... Σn= n(n+1)/2 Σ1 = n
use A P summation formula n/2(2a+(n-1)d)
sum of natural no is n*(n+1)/2 therefore that of 2n-1 is n(n+1)-1
Σn=(n)(n+1)/2 Σ(2n-1)=Σ(2n)-Σ(1) = 2Σn- n= 2n(n+1)/2 -n =n*n
Isn't that a little overcomplicated?
S_n = (n^2 - (n-1)^2) + ((n-1)^2 - (n-2)^2) + ... + (1 - 0)
The only terms in this sum that don't cancel are the first and last: n^2 and 0.
Partial sums arrive at the same answer but I think the above logic is what the question is contructed around.
Indeed :)
I also did it this way, telescoping sums for the win
Is there a need to calculate for Tn = 2n-1? As far as I can see the sequence resolves to Sn = n2 as all the other terms cancel out and (n-1)2 for t1 is zero. Wouldn't this give constant time solution? Or am I missing something?
formula of sum of AP series (which is 1,3,5,...,2*n-1) n/2*(a+(n-1)*d) given d=2 and a=1(as a first term and d is difference between two numbers in the series) which makes above equation n*n