A-Level Further Maths D4-01 Method of Differences: Example 1

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[@SIRMAPANDAONLINE]

Good work sir 👏

[@rapfarsibaza]

Thanks 👍

[@giorgixyz1909]

thank you !

[@asmartguycodename4721]

Is there any proof for taking the rth term to be a polynomial of degree (x+1) where x is the number of times we try to take the difference between consecutive terms to get an AP/ GP. please tell me.

[@TLMaths]

I don't know I'm afraid.

[@asmartguycodename4721]

@@TLMaths well, i hope to be in USA or any other European country, atleast i will be off from competition.

[@aghaahmed5946]

How were you able to deduce that the sums would be equal to each other from the first part?

[@TLMaths]

We showed that (2r+1)^2-(2r-1)^2=8r. So r = (1/8)*((2r+1)^2-(2r-1)^2) SUM( (2r+1)^2-(2r-1)^2 ) = SUM( 8r ) SUM( (1/8)*((2r+1)^2-(2r-1)^2) ) = SUM( r )

[@moodymonstrosity1423]

How'd we know that last terms would be n-2 , n-1 , n like how'd we know it's in increasing order and if the difference is of 1 or not

[@TLMaths]

In a series, the numbers always go up in 1s. So if the last term is n, then the previous term will be n-1, and the term before that will be n-2

[@EE-Spectrum]

Could one factorise (2r+1)^2-(2r-1)^2 as the difference of squares instead of expanding the brackets?

[@TLMaths]

I don't see why not

[@EE-Spectrum]

Thanks. I just thought it might be simpler to factorise instead of expanding the brackets. You are doing a great job.

[@nosir1479]

That was literally my immediate thought but when I did it, I found that expanding brackets would've been easier