how do i find the nth term of a recurring sequence

asgeorgia · **Posted:** Thu, 11 Oct 2007 12:09:51 UTC

given the recurring sequence say Un+1 = (3Un)^(1/2) and Un = 1
how do i find an expression for Un in terms of n. please help me

Soroban · **Posted:** Thu, 11 Oct 2007 13:17:53 UTC

Hello, asgeorgia!

Quote:

Given the recurring sequence: . $U_{n+1} \:= \:\left(3U_n)^{\frac{1}{2}}$ and $U_n \,=\, 1$

Find an expression for U_n

in terms of

My first impulse is to crank out the first few terms and look for a pattern.

$\begin{array}{ccccccc}U_1 & = & 1 &=& 3^0\\ U_2 & = & \left(3\cdot1)^{\frac{1}{2}} & = & 3^{\frac{1}{2}} \\ U_3 & = & \left(3\cdot3^{\frac{1}{2}} & = & \left(3^{\frac{3}{2}\right)^{\frac{1}{2}} & = & 3^{\frac{3}{4} \\ U_4 &=& \left(3\cdot3^{\frac{3}{4}\right)^{\frac{1}{2}} &=&\left(3^{\frac{7}{4}}\right)^{\frac{1}{2}} &=&3^{\frac{7}{8}} \\ U_5 &=&\left(3\cdot3^{\frac{7}{8}\right)^{\frac{1}{2}} &=&\left(3\frac{15}{8}\right)^{\frac{1}{2}} &=&3^{\frac{15}{16}} \\ \vdots \end{array}$

See the pattern?

We have 3 raised to a fractional power.
. . The denominator is a power-of-2: . $2^{n-1}$
. . The numerator is one less than the denominator: . $2^{n-1} - 1$

Therefore: . $U_n \;=\;3^{\left(\frac{2^{n-1}-1}{2^{n-1}\right)}$

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how do i find the nth term of a recurring sequence

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