For the same arithmetic sequence, Sₙ is the sum of 1/(aₖaₖ₊₁) from k=1 to n. Find n when Sₙ=10/21.

01 / Standard#Your turn

a_k=2k-1,\quad S_n=\sum_{k=1}^n\frac1{a_ka_{k+1}}

Official paper · jm01-2021 · II.4(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Split each denominator into two simple fractions.

Hint 2

Consecutive terms then cancel.

Worked solution

Use the difference between consecutive odd reciprocals.
$\frac1{(2k-1)(2k+1)}=\frac12\left(\frac1{2k-1}-\frac1{2k+1}\right)$
Telescope the sum and solve the rational equation.
$S_n=\frac12\left(1-\frac1{2n+1}\right)=\frac n{2n+1}$
The resulting n is a positive integer.
$n/(2n+1)=10/21\Rightarrow21n=20n+10\Rightarrow n=10$

n=10.

Checks and common pitfalls: Retain the factor 1/2 in the partial-fraction identity.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Use the difference between consecutive odd reciprocals.
$\frac1{(2k-1)(2k+1)}=\frac12\left(\frac1{2k-1}-\frac1{2k+1}\right)$
Telescope the sum and solve the rational equation.
$S_n=\frac12\left(1-\frac1{2n+1}\right)=\frac n{2n+1}$
The resulting n is a positive integer.
$n/(2n+1)=10/21\Rightarrow21n=20n+10\Rightarrow n=10$

Think first. Reveal a hint when the class is ready.

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