Question 822530
1st year
12*200=2400
2nd year
1.1*200=220
12*220=2640
3rd year
1.1*220=242
242*12=2904
4th year
1.1*242=266.20
12*266.20=3194.40
5th year 292.82
12*292.82=3513.84
2400+2640+2904+3194.40+3513.84=14652.24 total

another way
2400*1.1=2640
2640*1.1=2904
2904*1.1=3194.40
3194.40*1.1=3513.84
2400+2640+2904+3194.40+3513.84=14652.24 total
another way 
Tn = t1 * r^(n - 1)
find the nth number of a geometric sequence
where t1 is the starting amount
r is the ratio and 
n is the number of times/years
2400 * 1.1^(0)=2400
2400 * 1.1^(1)=2640
2400 * 1.1^(2)=2904
2400 * 1.1^(3)=3194.40
2400 * 1.1^(4)=3513.84
2400+2640+2904+3194.40+3513.84=14652.24 total

and finally the fastest way
the sum of  a geometric sequence
where t1 is the starting amount
r is the ratio and 
n is the number of times or years
S is the sum
S=t1*(1 - r^n)/(1 - r)
S=2400*(1 - 1.1^5)/(1 - 1.1)
S = 14652.24