If bob deposits $5,000 at the beginning of each year for 10 years in an account paying 12% interest compounded annually, find the amount he will have at the end of the 10 years. a. $98,272.92 c. $94,625.32 b. $98,828.28 d. $95,876.82

End of 10 years = 5,000 * (1.12)^10 =5,000*3.11 =  15,529.24
end of 9 years = 5,000 * (1.12)^9 = 5,000*2.77 = 13,865.39
end of 8 years = 5,000 * (1.12)^8 =5,000*2.48 = 12,379.82
end of 7 years = 5,000 * (1.12)^7 =5,000*2.21 = 11,053.41
end of 6 years = 5,000 * (1.12)^6 =5,000*1.97 = 9,869.11
end of 5 years = 5,000 * (1.12)^5 =5,000*1.76 = 8,811.71
end of 4 years = 5,000 * (1.12)^4 =5,000*1.57 = 7,867.60
end of 3 years = 5,000 * (1.12)^3 =5,000*1.40 = 7,024.64
end of 2 years = 5,000 * (1.12)^2 =5,000*1.25 = 6,272.00
end of 1 year = 5,000 * (1.12)^1 =5,000*1.12 = 5,600.00

TOTAL = 98,272.92

Answer is "a"

an easier way is to use the annuity formula (see attached)

Total = amount * [(1+r)^(n+1) -1] / r - amount
Total = 5,000 * [(1.12)^11] -1 / .12 -5,000
Total = 5,000 *[ 3.4785499933 -1] / .12  -amount
Total = (5,000 * 2.4785499933) / .12 -amount
Total = (5,000 * 20.6545832779) -amount
Total = 103,272.92 - 5,000.00
Total = 98,272.92

and here is an annuity calculator
http://www.1728.org/annuity.htm

Ais right if u want more just write the common factors (the numbers that you multiply)