Question

@mathmale quick question :)

Answer 1

is my answer correct and sufficient ?

Answer 2

they say the limit is 10 times, not an infinite number
so you should make the sum from n=1 to 10

Answer 3

put r=3/10
For the first change, she gets
35r
for the second change, she gets
35r^2
...
for the 10th change, she gets
35r^10. (not 11)

Total
=35r+35r^2+...35r^10
=35r(1+r+r^2+...+r^9)
=35r(r^10-1)/(r-1) by factoring r^n-1=(r-1)(1+r+r^2+...r^(n-1)

Answer 4

@phi do you think that's the only mistake i have? because mathmate explantations aren't clear with all these symbols.?

Answer 5

@skullpatrol help

Answer 6

I edited the symbols if it helps.
\(\color{navy}{\text{Originally Posted by}}\) \(\color{seagreen}{\text{mathmate}}\)
put r=3/10
For the first change, she gets
35r
for the second change, she gets
35r\(^{2}\)
...
for the 10th change, she gets
35r\(^{10}\). (not 11)

Total
=35r+35r\(^{2}\)+...35r\(^{10}\)
=35r(1+r+r\(^2\)+...+r\(^9\))
=\(\Large{\frac{35r(r^{10}-1)}{r-1}}\) by factoring r\(^n\)-1=(r-1)(1+r+r\(^2\)+...r\(^{(n-1)})\)
\(\color{navy}{\text{End of Quote}}\)

Answer 7

Unless that r^10-1 was supposed to be \(\Large{r^{10-1}}\) oops

Answer 8

@kittiwitti1 can you check another one? if yes post it here or another post?

Answer 9

Sorry, busy with multiple subjects