Question 8 of 12

Question 8 (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) The arithmetic set $A=\{2,5,8,\ldots,3n-1\}$ ($n$ terms, first term 2, common difference 3). (b) The recurrence $a_1=1$, $a_i=a_{i-1}+2i-1$ for $i\ge 2$.

Find. (a) A derivation of the closed-form sum $S$. (b) A closed form for $a_n$.

Approach. (a) apply the arithmetic-series sum formula directly. (b) unroll the recurrence into a sum of consecutive odd numbers and use the known odd-number sum identity.

1. (a) Sum of $A=\{2,5,8,\ldots,3n-1\}$. This is an arithmetic sequence with first term $a=2$, common difference $d=3$, and $n$ terms (the $k$-th term is $2+3(k-1)=3k-1$, matching the last term $3n-1$ at $k=n$). The arithmetic series sum formula is $S=\dfrac{n}{2}\big(2a+(n-1)d\big)$: $$S=\frac{n}{2}\big(2(2)+(n-1)(3)\big)=\frac{n}{2}\big(4+3n-3\big)=\frac{n}{2}(3n+1)$$ $$\boxed{S=\frac{n(3n+1)}{2}}$$
2. (b) Closed form for $a_n$. Unroll the recurrence from $i=2$ up to $i=n$ and telescope: $$a_n=a_1+\sum_{i=2}^{n}(2i-1)=1+\sum_{i=2}^{n}(2i-1)=\sum_{i=1}^{n}(2i-1)$$ since the $i=1$ term of $\sum(2i-1)$ is $2(1)-1=1=a_1$. The sum of the first $n$ odd numbers is the classical identity $\sum_{i=1}^n(2i-1)=n^2$ (each odd number $2i-1$ is the "gnomon" added to complete an $i\times i$ square from an $(i-1)\times(i-1)$ one): $$\boxed{a_n=n^2}$$ Check: $a_1=1=1^2$, $a_2=1+3=4=2^2$, $a_3=4+5=9=3^2$ — consistent with the recurrence.