2026 EJC P1 Q9

Junior College 2

11 marks

Let $U_1,U_2,U_3,\ldots$ denote the terms of a sequence and, for each $n$, $W_n=U_1+\cdots+U_n$, $n\geq1$.
It is given that $W_n=\frac{k^n}{3^{n-1}}-3$, where $k$ is a non-zero constant, $k\neq3$.
1. Find an expression for $U_n$ in terms of $k$ and $n$, and prove that the sequence is geometric.[4]
2. Determine the range of values of $k$ such that $W$, the sum to infinity of the series $W_n$, exists.[1]
A sequence $u_1,u_2,u_3,\ldots$ is such that $u_r=\frac1{r!}$, $r\geq1$.
1. Show, by writing out all the terms of the summation, that $\sum_{r=1}^{4}(u_r-u_{r+1})=u_1-u_5$.[1]
  It can be shown that $u_r-u_{r+1}=\frac1{r!+(r-1)!}$, for $r\geq1$.
2. Hence show that $\sum_{r=1}^{N}\frac1{r!+(r-1)!}=1-\frac1{(N+1)!}$.[2]
3. Hence show that $\frac1{2!+1!}+\frac1{3!+2!}+\frac1{4!+3!}+\cdots=\frac12$.[3]

Solution:

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Answer:(a)(i) $U_n=(k-3)(k/3)^{n-1}$ (a)(ii) Intended $\sum U_n$: $-3<k<0$ or $0<k<3$, sum $-3$. Literal $\sum W_n$: no allowed $k$ (b)(iii) $\frac12$

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