Expert: Josh - 1/5/2011

Question
1.find the sum to n terms of 8,88,888,8888......
2.〖0.2〗^log_√5⁡(1/4+1/8+1/16+⋯∞)

Answer
Hi Jazz,

Tricky questions. Here are the answers.
I don't know what comments are appropriate to add, as you did not mention any specific difficulties you have. I'd be happy to elaborate if you require further explanation. Feel free asking a follow-up. Here we go.

Q1. I formulated this problem using a first order recurrence equation. You will find eventually that there is a finite geometric series embedded in another geometric series.

Let S(n)=8+88+888+8888+... for n terms
Define T(0)=8. Subsequent terms are given by T(n)=T(n-1)+8*10^n, for n>=1  ....[#1]
Exercise: Verify yourself that this is true.

Goal: To find S(n)=T(0)+T(1)+T(2)+...+T(n-1).  This contains exactly "n" terms.
Using summation notation,

S(n)=sum_i T(i), for i=0:n-1 [this means index i goes from 0 to n-1, inclusive]
   =T(0) + sum_i T(i), for i=1:n-1 [now substituting expression [#1] in the series]
   =T(0) + sum_i [T(i-1)+8*10^i], for i=1:n-1 [by recursion, using [#1] again]
   =T(0) + sum_i [T(i-2)+8*10^(i-1)+8*10^i], for i=1:n-1 [repeating once more to make obvious]
   =T(0) + sum_i [T(i-3)+8*10^(i-2)+8*10^(i-1)+8*10^i], for i=1:n-1
By induction, we have
S(n)=T(0) + sum_i [T(0)+8*10+...+8*10^(i-2)+8*10^(i-1)+8*10^i], for i=1:n-1  ...[#2]
Note carefully that the expression inside the square bracket is subject to summation over i.

Now, using the formula for a finite geometric series:
a+a*r+a*r^2+...+a^(k-1) = a*(r^k-1)/(r-1) for r>1
We identify the first term a=80, r=10 and there are k=i terms.
Thus, the part "8*10+...+8*10^(i-2)+8*10^(i-1)+8*10^i" is replaced with 80(10^i-1)/(10-1).
So, [#2] becomes
S(n)=T(0) + sum_i[T(0)+(80/9)*(10^i-1)] ...[splitting the summation, isolating dependency on i]
   =T(0) + sum_i[T(0)-(80/9)] + (80/9)*sum_i[10^i], for i=1:n-1 [apply geometric series formula]
   =T(0) + sum_i[T(0)-(80/9)] + (80/9)*10(10^(n-1)-1)/(10-1), for i=1:n-1
   =T(0) + (n-1)*[T(0)-(80/9)] + (800/81)(10^(n-1)-1), for i=1:n-1
*** This is the final expression ***

Try n=3, for instance, [remembering that the first term T(0)=8]
S(3)=8+(3-1)*(8-(80/9))+(800/81)*(10^(3-1)-1)
   =984
   =888+88+8.

Q2. Let z=0.2^{log_sqrt(5)(S)}, where S=1/4+1/8+1/16+...
   Using the G.S. formula for an infinite sum,
   S=lim_{n->infinity} (1/4)*(1-(1/2)^n)/(1-(1/2))=(1/4)(1-0)/(1-(1/2))=1/2
   Hence, z=(1/5)^log_sqrt(5)(1/2) ...[#3]
   Using the change of base law, viz., log_x(y)= log_a(y)/log_a(x) for any a,
   we may rewrite log_{sqrt(5)}(1/2) as log_5(1/2)/log_5(sqrt(5)). This simplifies to
   log_5(1/2)/(1/2) since sqrt(5)=5^(1/2).
   So, [#3] becomes
   z=(1/5)^[log_5(1/2)/(0.5)]...[reciprocal of 0.5 is 2]
    =(1/5)^[2*log_5(1/2)]    ...[next, use x*log_a(y)=log_a(y^x)]
    =(1/5)^log_5(1/4)        ...[final trick is to recognize log_a(1/y)=log_{1/a}(y)]
    =(1/5)^log_{1/5}(4)
    =4

In the last line, recall that c=log_a(b), by definition, gives c as the exponent of a, such that a^c=b.