Double checking Proof of Summation Answer. (500 + 100 + 20 + ... + 4[5^(4 -n)] = 625( 1 - 1/5^n))

[jddoxtator]

The question:

Prove the statement is true for all natural numbers.

500 + 100 + 20 + ... + 4[5^(4 -n)] = 625( 1 - 1/5^n)

My answer:

Step 1:

For n=1,
4[5^(4 - 1)] = 625( 1 - 1/5^1)
4(125) = 625( 4/5)
500 = 500

The statement is true for n=1.

Step 2:

Assume n=k,
500 + 100 + 20 + ... + 4[5^(4 - k) = 625( 1 - 1/5^k)

Step 3:

Prove for (k + 1),
500 + 100 + 20 + ... + 4[5^(4 - k) + 4[5^[4 - (k + 1)]] = 625( 1 - 1/5^k) + 4[5^[4 - (k + 1)]] - Expression with factor including (k + 1) added to both sides.
= 625(1 - 1/5^k) + 4[5^4/5^(k +1)] - Negative exponent of 5 moved to denominator.
= 625(1 - 1/5^k) + 2500/5^(k + 1) - Simplify
= 625[(1 -1/5^k) + 4/5^(k + 1)] - Distributive property
= 625[1 - 5/5^(k + 1) + 4/5^ (k + 1)] - Fraction 1/5^k multiplied by 5 in both numerator and denominator to equalize bases to 5^(k + 1)
= 625[1 - 1/5^(k + 1)] - Simplify

The last statement is the statement to be proved for n = (k + 1)
This proves that the statement is true for all natural numbers n.

 

[Dr.Peterson]

Looks good to me.

Of course, some people would word parts of it differently; the only specific thing I'd want to see changed is that in step 2, you are not just "assuming n=k", but "assuming the claim is true for n=k". But that's just semantics.

Note that the key idea in writing step 3 is to have the last line in mind throughout, and do algebra to change the expression in that direction. This is much like proving a trig identity. And you did that well.

 

[lookagain]

Prove the statement is true for all natural numbers.
500 + 100 + 20 + ... + 4[5^(4 -n)] = 625( 1 - 1/5^n)

jddoxtator, you already worked out a solution, and you got a confirmation. Let me show an optional method.

Suppose \(\displaystyle \ S \ \) represents the sum of the finite series on the left-hand side.

\(\displaystyle S \ = \ 500 \ + \ 100 \ + \ 20 \ + \ ... \ + \ 4\cdot 5^{4 - n}\)

The next line reveals more terms on the right-hand side for the emphasis of previous exponents not shown before:

\(\displaystyle S \ = \ 500 \ + \ 100 \ + \ 20 \ + \ ... \ + \ 4\cdot 5^{6 - n} \ + \ 4\cdot 5^{5 - n} \ + \ 4\cdot 5^{4 - n}\)

\(\displaystyle S \ - \ 500 \ = \ 100 \ + \ 20 \ + \ ... \ + \ 4\cdot 5^{6 - n} \ + \ 4\cdot 5^{5 - n} \ + \ 4\cdot 5^{4 - n}\)

\(\displaystyle 5(S \ - \ 500) \ = \ 5(100 \ + \ 20 \ + \ ... \ + \ 4\cdot 5^{6 - n} \ + \ 4\cdot 5^{5 - n} \ + \ 4\cdot 5^{4 - n})\)

\(\displaystyle 5S \ - \ 2500 \ = \ 500 \ + \ 100 \ + \ ... \ + \ 4\cdot 5^{7 - n} \ + \ 4\cdot 5^{6 - n} \ + \ 4\cdot 5^{5 - n}\)

Add \(\displaystyle \ 4\cdot 5^{4 - n} \ \) to each side to create \(\displaystyle \ S \ \) on the right-hand side:

\(\displaystyle 5S \ - \ 2500 \ + \ 4\cdot 5^{4 - n} \ = \ 500 \ + \ 100 \ + \ ... \ + \ 4\cdot 5^{7 - n} \ + \ 4\cdot 5^{6 - n} \ + \ 4\cdot 5^{5 - n} \ + \ 4\cdot 5^{4 - n} \)

\(\displaystyle 5S \ - \ 2500 \ + \ 4\cdot 5^{4 - n} \ = \ 500 \ + \ 100 \ + \ 20 \ + \ ... \ + \ 4\cdot 5^{4 - n} \)

\(\displaystyle 5S \ - \ 2500 \ + \ 4\cdot 5^{4 - n} \ = \ S \)

\(\displaystyle 5S \ - \ S \ = \ 2500 \ - \ 4\cdot 5^{4 - n} \)

\(\displaystyle \dfrac{4S}{4} \ = \ \dfrac{2500}{4} \ - \ \dfrac{4\cdot 5^{4 - n}}{4} \)

\(\displaystyle S \ = \ 625 \ - \ 5^{4 - n}\)

\(\displaystyle S \ = \ 625 \ - \ \dfrac{5^4}{5^n}\)

\(\displaystyle S \ = \ 625 \ - \ \dfrac{625}{5^n}\)

\(\displaystyle S \ = \ 625\bigg(1 \ - \ \dfrac{1}{5^n}\bigg)\)