Trigonometry & Inverse Trigonometry Question - None

Question

28. If the equation $\sum_{n=0}^{10} \text{arc cot}\left(\frac{1+2^{2n+1}}{2^n}\right) = \text{arc cot}\frac{a}{b}$, where $a$ and $b$ are coprime positive integers. The value of $\log_2\left(\frac{b+a}{a-b}\right)$, is:

Accepted Answer

Step 1: Identify the telescoping identity

Use the identity: arc cot(A) - arc cot(B) = arc cot((AB+1)/(B-A))

Rearranging: arc cot((AB+1)/(B-A)) = arc cot(A) - arc cot(B)

Step 2: Express the general term in telescoping form

For the term arc cot((1+2^(2n+1))/2^n), let A = 2^n and B = 2^(n+1):

Then AB + 1 = 2^n · 2^(n+1) + 1 = 2^(2n+1) + 1 ✓

And B - A = 2^(n+1) - 2^n = 2^n(2-1) = 2^n ✓

Therefore: arc cot((1+2^(2n+1))/2^n) = arc cot(2^n) - arc cot(2^(n+1))

Step 3: Sum the telescoping series

∑(n=0 to 10) [arc cot(2^n) - arc cot(2^(n+1))]

= [arc cot(2^0) - arc cot(2^1)] + [arc cot(2^1) - arc cot(2^2)] + ... + [arc cot(2^10) - arc cot(2^11)]

= arc cot(1) - arc cot(2^11)

= arc cot(2048/2047)

Step 4: Identify a and b

Since arc cot(1) - arc cot(2^11) = arc cot((1·2^11 + 1)/(2^11 - 1)) = arc cot(2049/2047)

Therefore: a = 2049, b = 2047 (these are coprime)

Step 5: Calculate the logarithm

log₂((b+a)/(a-b)) = log₂((2047+2049)/(2049-2047))

= log₂(4096/2)

= log₂(2048)

= log₂(2^11)

= 11

∴ Answer: 11

Question:

Step-by-Step Solution

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