Thanks for you really helpful site! I have a question: Why is the sum of 2 not just 2? Surely 2 doesn’t vary as r goes up… It’s probably an obvious answer but I just can’t get my head around it.
Everything on the right of the Σ gets added up for each value of the argument. For example, Σf(r) for r = 0 to 3 is f(0) + f(1) + f(2) + f(3). If f(r) = (2 + 4^r), then Σf(r) for r = 0 to 3 is (2+4^0) + (2+4^1) + (2+4^2) + (2+4^3).
That’s how you end up with 8 2’s instead of just one.
Thank you!! Your site has really helped me a lot!
It’s a pleasure — glad to hear it’s helped.
I don’t see where the -4 in the second line is coming from, could you explain that?
Hi Jonas, thanks for your question.
If you look at the sum S, it expands into the values:
4 + 16 + 64 + 256 + 1024 + 4096 + 16384 + 65536
If you consider 4S, you would get this by multiplying each value of S by 4, giving:
16 + 64 + 256 + 1024 + 4096 + 16384 + 65536 + 4 x 65536
Notice that the the seven terms 16 + 64 + 256 + 1024 + 4096 + 16384 + 65536 are shared between S and 4S. In effect, multiplying S by 4 just ‘shifts the terms of S to the right’ by one: so 4S is S with the leftmost number (4) removed and a new number (4 x 65536)added on to the right . You should be able to see from this that 4S = S + 4 x 65536 – 4.
Hi Rachida, thanks for your question.
The formula that you have used for the sum of a GP is for the sum from i = 0 , whereas the question here asks for the sum from i=1. So you need to be careful when using such formulae. I prefer to use the approach above because you only have to remember one thing (to sum a GP, multiply both sides by the common ration and solve for the sum), rather than three things: the formula, whether it is for the sum from i=0 or i=1, whether it assumes the last term in the sum is i=n or i=n-1.
Doesn’t the sum we have here of 4^r if a sum of a geometric suite, so we could calculate the Sigma from 1 to 8 of 4^r = ( 1- 4^8)/(1-4) like this?
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