For the preceding sequence, sum bₙ=1/Sₙ to obtain Tₙ.

01 / Standard#Your turn

Official paper · jm01-2026 · II.3(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

First calculate Sₙ.

Hint 2

Split 1/[n(n+2)] into two simple fractions.

Worked solution

Find and decompose the reciprocal partial sum.
$S_n=\frac n2(12+8n+4)=4n(n+2),\quad b_n=\frac18\left(\frac1n-\frac1{n+2}\right)$
All interior terms cancel, leaving two initial and two terminal terms.
$T_n=\frac18\left(1+\frac12-\frac1{n+1}-\frac1{n+2}\right)=\frac3{16}-\frac18\left(\frac1{n+1}+\frac1{n+2}\right)$

Tₙ=3/16−(1/8)(1/(n+1)+1/(n+2)).

Checks and common pitfalls: A gap of two in the denominators leaves two terminal terms.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Find and decompose the reciprocal partial sum.
$S_n=\frac n2(12+8n+4)=4n(n+2),\quad b_n=\frac18\left(\frac1n-\frac1{n+2}\right)$
All interior terms cancel, leaving two initial and two terminal terms.
$T_n=\frac18\left(1+\frac12-\frac1{n+1}-\frac1{n+2}\right)=\frac3{16}-\frac18\left(\frac1{n+1}+\frac1{n+2}\right)$

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

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