On the first of each month Mr. Tripathy deposits Rs. 1000 in a savings account that pays 3% compounded monthly. Assuming that no withdrawals are made, the recurrence relation for the total amount of money in the account at the end of n months...........
Answers
Given : On the first of each month Mr. Tripathy deposits Rs. 1000 in a savings account that pays 3% compounded monthly.
no withdrawals are made
To Find : total amount of money in the account at the end of n months.
Solution:
Amount Deposited Each Month = n
A = P(1 + R/100)ⁿ
P = 1000
R = 3 %
Amount Deposited in 1st month become after n month = 1000( 1 + 3/100)ⁿ
= 1000( 1.03)ⁿ
Amount deposited 2nd month = 1000(1.03)ⁿ⁻¹
and so on
Amount deposited in nth month = 1000(1.03)¹
This forrms an Gp
1000( 1.03)ⁿ , 1000(1.03)ⁿ⁻¹ , ,,,,,,,,,,,,,,,,,, 1000(1.03)¹
= 1000(1.03)¹ , ..................1000(1.03)ⁿ⁻¹ , 1000( 1.03)ⁿ
r = n
a = 1000(1.03)
Sum of GP = a(rⁿ - 1)/(r -1 )
= 1000(1.03) (1.03ⁿ - 1)/(1.03 - 1)
= 34,333.33 (1.03ⁿ - 1)
total amount of money in the account at the end of n months = 34,333.33 (1.03ⁿ - 1)
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