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i suppose not being in the tags means i still lack in notoriety huh

what?

nope, they help me a lot

you have those sum formulas ?

and did you get that long explanation ? :P

why a1=1?

oh, it seems easy. Let me try

but for part a.
100*(1* 0.2 ^(8) ) ???

uhh.

0.000256???!

OMG, there must be some mistake in my method....it can't be 0.0256
let me review it...

430?

yes, i get 430 too :)

thanks. and part b Emily?

revised formula for sum
\(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)

a1 = 100, r =0.2, n =9
you'll get the answer :)

i know why i cannot get the answer. i think n=8...
thankd

thanks*

after 1 year means n=2, (100*0.2 is the 2nd term)
so, after 8 years, means n=9

ah ha, i got it.

not total amount, IN THAT YEAR

okay..

so, you will use n'th term formula or sum formula ?

n'th term

correct!
and since its \(\large 10th\) year, \(\large n=11\)

Emily=$619.174
Bobby=$600
then Emily's claim is correct, right?

and part c ii ?
Bobby=$7700?
Emily=$??

ohh...for part 2 we need sum formula.....

yes

Emily : 100 * [(1.2)^11 -1] / 0.2 =... ?
i plugged in, r=0.2,a1 = 100, n=11

3215...

is that the answer in your book ?
difference = 7700-3215

the answer is 635

i am getting 3850 for bobby

3850-3215=635

Sum for bobby
\(\large (11/2)[2*100+10*50]=3850\)

why 2 times 100?

the formula \(\large (n/2) [\huge {\color{red} 2}a_1+(n-1)d]\)

uh? why? it's fixed

yeah, thats the general formula for sum...

uh, OMG .....
okay, let me see. how about Emily's?

sum formula i posted above, the "revised" one
Emily : 100 * [(1.2)^11 -1] / 0.2

\(S_n = a_1 [\dfrac{(r+1)^n-1}{r}]\)

ah ha, thanks

3215...
thank you so much

welcome ^_^