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A sequence is given by $U_m =2U_n - 1$ where m is $n+1$. $U_1=2$ for $n\geq1$. Find the term of the sequence that has value 257.

Approach:

$U_m$=$2U_n - 1 = 257$

$U_n = 129$

$U_1 = 2$

$U_2 = 3$

$U_3 = 5$

$U_4 = 9$

$U_5 = 17$

Of course, if I continue to find the sequence for $U_n$, i will eventually found $U_8$ to be 129 and I can continue to find $U_m$ easily which is $U_9$.

I would like to know how can I find the answer more easily besides doing this, especially the value of the term gets even larger?

Joe
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  • One technique is to take the differences of the numbers and see if you see a pattern. (which is basically the first approach in answering any "what's the next number in the sequence" problem) – Mike O'Connor Jul 13 '22 at 03:36
  • Hint: $;U_{n+1} \color{red}{-1} =2U_n - 1 \color{red}{-1},$. – dxiv Jul 13 '22 at 03:41

2 Answers2

4

Hint: see if you can spot a pattern in the sequence $U_n -1$

jjagmath
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1

Let $$ d_n=U_{n+1}-U_{n} \quad \forall n\ge 1. $$ Then $d_1=U_2-U_1=2U_1-1-U_1=1$ and for all $n\ge 1$ we have $$ d_{n+1}=U_{n+2}-U_{n+1}=2U_{n+1}-2U_{n}=2d_n, $$ i.e. $d_n$ is a geometric sequence. Therefore $$ d_n=2^{n-1}d_1=2^{n-1}. $$ Finally we have \begin{eqnarray} U_n&=&(U_n-U_{n-1})+(U_{n-1}-U_{n-2})+\cdots+(U_3-U_2)+(U_2-U_1)+U_1\cr &=& U_1+\sum_{i=1}^{n-1}(U_{i+1}-U_{i})\cr &=&2+\sum_{i=1}^{n-1}d_i\cr &=&2+\sum_{i=1}^{n-1}2^{i-1}\cr &=&2+\sum_{k=0}^{n-2}2^k\cr &=&2+\frac{1-2^{n-2+1}}{1-2}\cr &=&2-(1-2^{n-1})\cr \end{eqnarray} Hence $$ U_n=1+2^{n-1} \quad \forall n\ge 1. $$ Notice that with this formula, we have \begin{eqnarray} U_1&=&2\cr U_2&=&3\cr &\vdots&\cr U_8&=&129\cr U_9&=&257\cr U_{10}&=&513\cr U_{11}&=&1025 \end{eqnarray}

HorizonsMaths
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