Last ones I need help with!
so the common difference =d= +2 a1 = January clients = 5 use the formula \(\Large a_n = a_1 +(n-1)d\)
december means n = 12
Alright so a_n=a_1(n-1)d a_n=5+(n-1)2 a_n=5+(12-1)2 a_n=5+11(2) a_n=5+22 a_n=27
correct, 27 clients in december
to get the total number of clients in 12 months, we need to add all the clients she added each month! we could use this sum formula \(\Large S_n = (n/2) (2a+(n-1)d)\) n = 12
that \(a\) is actually \(a_1\)
S_n=(n/2)(2a_1+(n-1)d) S_n=(12/2)(2(5)+(12-1)2) S_n=(6)(10+(11)2) S_n=(6)(10 + 22) S_n=(6)(32) S_n=192
correct :) ask if any doubts...
Hahaa. Okay great so I don't know how to do the sigma ones at all.
\(\Large \sum = sum\) \(\sum \limits_{n=1}^6\) means sum from 1st term to 6th term
one easier approach to do both 2 and 3, (just because the upper limit is small) is to keep on plugging in numbers for n plug in n =1, then n=2, then n=3....n=6 and add all the answers
for 3. start with n = 0, go till n =5 add all the answers
So for 2 n^2+1 1^2+1=1+1=2 2^2+1=4+1=5 3^2+1=9+1=10 4^2+1=16+1=17 5^2+1=25+1=26 6^2+1=36+1=37 2+5+10+17+26+37=97
too lazy to check, i assume thats correct :P
yup, 97 is correct
Lol okay For 3 16(1/4)^n 16(1/4)^0=16*1=16 16(1/4)^1=16*.25=4 16(1/4)^2=16*.0625=1 16(1/4)^3=16*.015625=.25 16(1/4)^4=16*.00390625=.0625 16(1/4)^5=16*9.765625E-4=.015625 16+4+1+.25+.0625+.015625=21.328125
correct! :)
Ok so how do I determine the last one.
can you find the common ratio between the terms of the last sequence ?
if magnitude of common ratio |r| > 1 , then the series does not converge at all and the sum would be infinite
So to find r Do I do it like this? 1/4 to 1/2 or is it 1/4 to -1/2?
r = 2nd term / 1st term = 3rd term/ 2nd term = ...
-1/2 divided by 1/4 or simply -2 divided by 1
-1/2 divided by 1/4 = -2 -2 divided by 1= -2 |r|>1 |-2|>1 2>1
correct, so the sum does not even exist, as the sequence is divergent
Alright thank you!
welcome ^_^
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