Sum cₙ=aₙbₙ for the preceding two sequences.

01 / Standard#Your turn

c_n=2(2n+1)3^{n-1}

Official paper · jm01-2025 · II.1(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Look for consecutive differences involving n3ⁿ.

Hint 2

Compare 2n3ⁿ with 2(n−1)3ⁿ⁻¹.

Worked solution

Rewrite each summand as a difference.
$c_k=2k3^k-2(k-1)3^{k-1}$
Sum from k=1 to n; all internal terms cancel.
$T_n=\sum_{k=1}^n[2k3^k-2(k-1)3^{k-1}]=2n3^n$

Tₙ=2n·3ⁿ.

Checks and common pitfalls: A product sequence is not obtained by multiplying the two partial sums.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Rewrite each summand as a difference.
$c_k=2k3^k-2(k-1)3^{k-1}$
Sum from k=1 to n; all internal terms cancel.
$T_n=\sum_{k=1}^n[2k3^k-2(k-1)3^{k-1}]=2n3^n$

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

2025 JM01