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- Feb 14, 2012

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Prove that $|S_n-3T_n|=2$.

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- Thread starter
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- #1

- Feb 14, 2012

- 3,909

Prove that $|S_n-3T_n|=2$.

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\(\displaystyle S_n=\sum_{k=0}^{3n}{3n \choose k}=(1+1)^{3n}=8^n\)

For the second sum, I looked at the first 5 values:

\(\displaystyle T_1=2,\,T_2=22,\,T_3=170,\,T_4=1366,\,T_5=10922\)

and determined the recursion:

\(\displaystyle T_{n+1}=7T_{n}+8T_{n-1}\)

The characteristic equation for this recursion is:

\(\displaystyle r^2-7r-8=(r+1)(r-8)=0\)

and so the closed form is:

\(\displaystyle T_{n}=k_1(-1)^n+k_28^n\)

Using the initial values to determine the parameters, we may write:

\(\displaystyle T_{1}=-k_1+8k_2=2\)

\(\displaystyle T_{2}=k_1+64k_2=22\)

Adding the two equations, we find:

\(\displaystyle 72k_2=24\implies k_2=\frac{1}{3}\implies k_1=\frac{2}{3}\)

Hence:

\(\displaystyle T_{n}=\frac{1}{3}\left(2(-1)^n+8^n \right)\)

And so we find:

\(\displaystyle \left|S_n-3T_n \right|=\left|8^n-3\left(\frac{1}{3}\left(2(-1)^n+8^n \right) \right) \right|=\left|8^n-2(-1)^n-8^n \right|=\left|2(-1)^{n+1} \right|=2\)

Shown as desired.

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- Feb 14, 2012

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Hey thanks for participating

\(\displaystyle S_n=\sum_{k=0}^{3n}{3n \choose k}=(1+1)^{3n}=8^n\)

For the second sum, I looked at the first 5 values:

\(\displaystyle T_1=2,\,T_2=22,\,T_3=170,\,T_4=1366,\,T_5=10922\)

and determined the recursion:

\(\displaystyle T_{n+1}=7T_{n}+8T_{n-1}\)

The characteristic equation for this recursion is:

\(\displaystyle r^2-7r-8=(r+1)(r-8)=0\)

and so the closed form is:

\(\displaystyle T_{n}=k_1(-1)^n+k_28^n\)

Using the initial values to determine the parameters, we may write:

\(\displaystyle T_{1}=-k_1+8k_2=2\)

\(\displaystyle T_{2}=k_1+64k_2=22\)

Adding the two equations, we find:

\(\displaystyle 72k_2=24\implies k_2=\frac{1}{3}\implies k_1=\frac{2}{3}\)

Hence:

\(\displaystyle T_{n}=\frac{1}{3}\left(2(-1)^n+8^n \right)\)

And so we find:

\(\displaystyle \left|S_n-3T_n \right|=\left|8^n-3\left(\frac{1}{3}\left(2(-1)^n+8^n \right) \right) \right|=\left|8^n-2(-1)^n-8^n \right|=\left|2(-1)^{n+1} \right|=2\)

Shown as desired.

- Mar 10, 2012

- 835

Define $f(x)=\sum_{k=0}^{3n}{3n\choose k}x^k$.

Prove that $|S_n-3T_n|=2$.

Then $f(1)+f(\omega)+f(\omega^2)=3T_n$, where $\omega$ is a complex cube root of unity. Note that $f(1)=S_n$. So we get $S_n-3T_n=-[(1+\omega)^{3n}+(1+\omega^2)^{3n}]=-2$

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- Feb 14, 2012

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HiDefine $f(x)=\sum_{k=0}^{3n}{3n\choose k}x^k$.

Then $f(1)+f(\omega)+f(\omega^2)=3T_n$, where $\omega$ is a complex cube root of unity. Note that $f(1)=S_n$. So we get $S_n-3T_n=-[(1+\omega)^{3n}+(1+\omega^2)^{3n}]=-2$

Thanks for participating and I really appreciate you adding another good solution to this problem and my thought to solve this problem revolved around the idea that you used too!