Challenge compound interest problem

[avishal]

Hi,

I've been impressed with the Manhattan challenge problem series, so here's my two cents. I came across this problem today while calculating the return for my investment :)

Question: I plan to buy an investment instrument for a 20 years term. The investment company guarantees 10% annual rate of return compounded yearly. If I invest $20,000 per year throughout the term period (i.e. 20 years), what's the total fund value at the end of term period?

Solution:

Compound Interest formula = P*((1 + r/100)^n), where P is principal, r is rate per annum, and n is number of years

So, basically, we need to find out the following:

20,000*((1+10/100)^20) + 20,000*((1+10/100)^19) + 20,000*((1+10/100)^18) + .. + 20,000*((1+10/100)^1)

= 20,000*[x^20 + x^19 + .. + x], where x = 1 + 10/100

Getting x^n + x^(n-1) + x^(n-2) + ... + x is the tricky part.

Say y = x^n + x^(n-1) + x^(n-2) + ... + x --- (1)

From 1,

y - x^n = x^(n-1) + x^(n-2) + ... + x

=> y - x^n + 1 = x^(n-1) + x^(n-2) + ... + x + 1---- (2)

Also, from 1,

y = x(x^(n-1) + x^(n-2) + ... + x + 1) --- (3)

Combining 2 and 3,

y = x ( y - x^n + 1)

=> y = xy - x^(n+1) + x

=> x^(n+1) - x = xy - y

=> y(x-1) = x(x^(n) - 1)

=> y = x(x^(n) - 1) / (x - 1)

check if formula is correct (example):

What is 3^4 + 3^3 + 3^2 + 3 ?

= 81 + 27 + 9 + 3
= 120

Using the formula (y = x(x^(n) - 1) / (x - 1)):

3^4 + 3^3 + 3^2 + 3 = 3(3^4 - 1) / (3 -1)

= 3(81 - 1) / (3 - 1)
= 3*80/2
= 3*40
= 120

Now, using the formula, after 20 years, the total fund value would be 20,000 * [1.1(1.1^20 - 1)/(1.1-1)]; x = 1 + 10/100 = 110/100 = 1.1

= 20,000 * [1.1 * 6.7275 / 0.1]
= 20,000 * 11 * 6.7275
= $1,480,050

Hope you find it useful.

Regards,
vishal

[shaji]

The math logic is perfect. A calculatio error in the penultimate step. Missed out on (1.1^20 -1) part. The correctanswer is $126005/-

Hi,

I've been impressed with the Manhattan challenge problem series, so here's my two cents. I came across this problem today while calculating the return for my investment :)

Question: I plan to buy an investment instrument for a 20 years term. The investment company guarantees 10% annual rate of return compounded yearly. If I invest $20,000 per year throughout the term period (i.e. 20 years), what's the total fund value at the end of term period?

Solution:

Compound Interest formula = P*((1 + r/100)^n), where P is principal, r is rate per annum, and n is number of years

So, basically, we need to find out the following:

20,000*((1+10/100)^20) + 20,000*((1+10/100)^19) + 20,000*((1+10/100)^18) + .. + 20,000*((1+10/100)^1)

= 20,000*[x^20 + x^19 + .. + x], where x = 1 + 10/100

Getting x^n + x^(n-1) + x^(n-2) + ... + x is the tricky part.

Say y = x^n + x^(n-1) + x^(n-2) + ... + x --- (1)

From 1,

y - x^n = x^(n-1) + x^(n-2) + ... + x

=> y - x^n + 1 = x^(n-1) + x^(n-2) + ... + x + 1---- (2)

Also, from 1,

y = x(x^(n-1) + x^(n-2) + ... + x + 1) --- (3)

Combining 2 and 3,

y = x ( y - x^n + 1)

=> y = xy - x^(n+1) + x

=> x^(n+1) - x = xy - y

=> y(x-1) = x(x^(n) - 1)

=> y = x(x^(n) - 1) / (x - 1)

check if formula is correct (example):

What is 3^4 + 3^3 + 3^2 + 3 ?

= 81 + 27 + 9 + 3
= 120

Using the formula (y = x(x^(n) - 1) / (x - 1)):

3^4 + 3^3 + 3^2 + 3 = 3(3^4 - 1) / (3 -1)

= 3(81 - 1) / (3 - 1)
= 3*80/2
= 3*40
= 120

Now, using the formula, after 20 years, the total fund value would be 20,000 * [1.1(1.1^20 - 1)/(1.1-1)]; x = 1 + 10/100 = 110/100 = 1.1

= 20,000 * [1.1 * 6.7275 / 0.1]
= 20,000 * 11 * 6.7275
= $1,480,050

Hope you find it useful.

Regards,
vishal

[StaceyKoprince]

Thanks, Vishal - just want to check something. Do you mean that you created this problem yourself or you found it somewhere? If you found it somewhere, please cite the source of the problem - we have to cite anything we do not write ourselves. If you made it yourself, thanks for the contribution!

[netcaesar]

Can anyone explain how do you combine 2 and 3?

From 1,

y - x^n = x^(n-1) + x^(n-2) + ... + x

=> y - x^n + 1 = x^(n-1) + x^(n-2) + ... + x + 1---- (2)

Also, from 1,

y = x(x^(n-1) + x^(n-2) + ... + x + 1) --- (3)

Combining 2 and 3,

y = x ( y - x^n + 1)

=> y = xy - x^(n+1) + x

=> x^(n+1) - x = xy - y

=> y(x-1) = x(x^(n) - 1)

=> y = x(x^(n) - 1) / (x - 1)

[avishal]

Thanks, Vishal - just want to check something. Do you mean that you created this problem yourself or you found it somewhere? If you found it somewhere, please cite the source of the problem - we have to cite anything we do not write ourselves. If you made it yourself, thanks for the contribution!

Really a very late response, but I created this myself when I was working on buying a unit-linked insurance policy for myself.

Regards,
Vishal

[JonathanSchneider]

Beautiful!

netcaesar, the combination from lines 2 and 3 is based on the fact that everything on the right-hand side of the equation in 2 is equal to the interior of the parentheses in 3.