AMC 10 · 2025 · #7

Grade 11 algebra

Archetype Telescoping Collapse └ Sub-archetype Difference-Split Collapse

logarithm-propertiestelescoping-sumfraction-arithmetic easier-related-problempattern-recognition ↑ Prerequisites: logarithm-properties

📏 Medium solution 💡 3 insights

Problem

Add up the 254 terms (log₂(1+1/n))/(log₂n)(log₂(n+1)), one for each whole number n from 2 to 255, and give the total exactly.

Pick an answer.

(A)

$\frac{3}{4}$

(B)

$1 -\frac{1}{\log_{2}255}$

(C)

$\frac{7}{8}$

(D)

$\frac{15}{16}$

(E)

How to solve

Strategy Organize Information in More Ways

Adding 254 terms is impossible by hand, so the term itself has to be rewritten into a friendlier shape. Tool #15 (Organize Information in More Ways) drives that: the numerator is one logarithm of a quotient, and the denominator holds the two logarithms log₂n and log₂(n+1) separately, so rewriting the numerator as a difference of exactly those two logarithms makes the fraction split apart. Tool #9 (Solve an Easier Related Problem) then tests the rewritten sum on just three terms, where the cancellation is visible by eye, and tool #5 (Look for a Pattern) extends that cancellation to all 254 terms. Tool #3 (Eliminate Possibilities) settles the finish, because the stopping point 255 was chosen so that 256=2⁸ appears, and two of the choices are near-misses that differ only in which logarithm survives at the end.

1STEP 1

Rewrite the numerator as a difference

Rewrite the numerator as a difference of logs.

log₂(1+1/n)=log₂ (n+1)/n=log₂(n+1)-log₂n

2STEP 2

Split the term into two reciprocals

Each term splits into a difference of reciprocals.

(B-A)/AB=1/A-1/B ⟹ (log₂(1+1/n))/(log₂n)(log₂(n+1))=1/log₂n-1/(log₂(n+1))

3STEP 3

Test a three-term version

A three-term version confirms the cancellation.

Σ_n=2⁴(1/log₂n-1/(log₂(n+1)))=1/log₂₂-1/log₂₅

4STEP 4

Cancel across the full sum

Only the two end terms survive.

Σ_n=2²⁵⁵(1/log₂n-1/(log₂(n+1)))=1/log₂₂-1/log₂₂₅₆

5STEP 5

Evaluate the two survivors

They are one and one eighth.

log₂₂=1, log₂₂₅₆=log₂₂⁸=8

6STEP 6

Subtract and pick the choice

Subtracting gives seven eighths.

1/log₂₂-1/log₂₂₅₆=1-1/8=7/8

Answer

7/8

Stopping the same telescoping at any upper limit N gives 1-1/(log₂(N+1)), which is always below 1 and creeps upward as N grows, so the total must be less than 1 and choice (E) is impossible. A floor check agrees: the single term at n=2 is already 1-1/log₂₃≈ 0.369, and the remaining 253 terms are all positive, so the total sits comfortably above 3/4. Note that estimating alone cannot finish the job, since choice (B) is 1-1/log₂₂₅₅≈ 0.87489 while 7/8=0.875; only the exact endpoint log₂₂₅₆=8 separates them, and the last term of the sum genuinely ends at n+1=256.

💡Key takeaway

When each term of a long sum can be written as one value minus the next value of the same expression, everything in the middle cancels and only the first piece minus the last piece is left.

Rewrite the numerator as a difference
Split the term into two reciprocals
Test a three-term version
Cancel across the full sum
Evaluate the two survivors
Subtract and pick the choice