#Series
43 messages · Page 1 of 1 (latest)
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well, for starters, lets split up the sum
$\sum_{r=1}^{12} 3r^2+\sum_{r=1}^{12} 8r+\sum_{r=1}^{12} 3+k\sum_{r=1}^{12} 2^{r-1}=3520$
;( | 追放された興奮
yupp i agree
im not sure how to do the one with the k with 2^r-1 but i do know the others
2^[r-1} is just a geometric series
but i dont understand what would i do with it?
$\sum_{k=1}^{n} ar^k= \frac{a(1-r^n)}{1-r}$
;( | 追放された興奮
2^{r-1}=2^r * 1/2
so where there is a r in the fraction on the right, do i sub in 1/2n(n+1)?
wdym
im not sure, its my first time seeing smthing like that
hmm
do u think i should just skip this question?
no, it is achievable
here, let me simplify
okii
$\sum_{r=1}^{12} \frac{2^r}{2}$
;( | 追放された興奮
ah apologies, i understand
im not completely sure how you manipulated 2^r-1 in that form
i used an exponent law
$a^{b-c}=\frac{a^b}{a^c}$
;( | 追放された興奮
wait i think ik how to do it now
i will be back in like another hour
nws, i think i can do the question 😄
tysm for ur help