How do I differentiate #50sum_(i=1)^x(1.03^i)#?

1 Answer
Steve M
Jan 18, 2018

# d/dx 50 \ sum_(i=1)^x \ (1.03)^i = (5150 \ ln 1.03)/3 \ 1.03^x #

# " " ~~ 50.742610... \ 1.03^x #

Explanation:

First let us consider the sum:

# S = sum_(i=1)^n \ (1.03)^i #

For clarity, let us expand the summation:

# S = 1.03 + 1.03^2 + 1.03^3 + ... + 1.03^n #

It should be clear that this summation is that of a GP (Geometric Progression) with first term #a=1.03# and common ratio #r=1.03#/ As such we can use the GP summation formula:

# S_n = (a(r^n-1))/(r-1) #

# \ \ \ \ = (1.03(1.03^n-1))/(1.03-1) #

# \ \ \ \ = (1.03)/(0.03)(1.03^n-1) #

# \ \ \ \ = (103)/(3)(1.03^n-1) #

Using this summation result we can now write the problem as:

# d/dx 50 \ sum_(i=1)^x \ (1.03)^i = d/dx { (50 xx 103)/(3)(1.03^x-1) }#

# " " = 5150/3 \ d/dx { 1.03^x-1 }#

# " " = 5150/3 \ d/dx { 1.03^x }#

# " " = 5150/3 \ d/dx { e^(ln 1.03^x) }#

# " " = 5150/3 (ln 1.03) e^(ln 1.03^x) #

# " " = (5150 \ ln 1.03)/3 \ 1.03^x #

# " " ~~ 50.742610... \ 1.03^x #

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