#how to do this series question
65 messages · Page 1 of 1 (latest)
its definition not a typo?
jus checking
bc usually the expression is always in terms of r
wait it cant even be a typo bc it would cancel out
lemme try the q and ill get back to you
treat 2n + 1 as a constant
nah just think about it as if its a constant
alr lemme finish
icl this question is cooked 💀
what i got doesnt factorise into the form
Ok lemme try stay there
Is this for ii
Yeah
I got an answer dont know if its right
Workings?
But you are summing up the 2n+1 bit n+1 times from r=0 to n
So its Σr² - 2Σr + (n+1)(2n+1) you can simplify that and factorise
Pffff I’m confused
Which bit
You can still use the standard results for r and r² since adding 0 doesnt change them
Oh I see
The 0 affects the constants
I get the r^2 and -2r but not next bit
5
Yes since there are 5 terms you do 5(5) = 25 in total
Yes
Don’t get
If its Σ(r+5) from r = 1 to n how many times do u add 5
5…
Ye so that summation can be rewritten as Σr + 5n
Ok yeah
From r=0 to n how many terms are there
Ye
Still can’t do it lol
Ye im redoing it and it wont factorise lol
Ye i got the same