Author Topic: Harder induction (Read 1122 times) Tweet Share

hanaacdr · « **on:** January 08, 2017, 12:02:48 pm »

Hi could i please get some help on this question:

If Un = 2^(n+2) + 3^(2n+1)
show that Un+1= 2Un + 7.3^(2n+1)
and hence show that Un is divisible by 7 for n>= 1

thank you

RuiAce · « **Reply #1 on:** January 08, 2017, 12:09:12 pm »

Quote from: hanaacdr on January 08, 2017, 12:02:48 pm

Hi could i please get some help on this question:

If Un = 2^(n+2) + 3^(2n+1)
show that Un+1= 2Un + 7.3^(2n+1)
and hence show that Un is divisible by 7 for n>= 1

thank you

Remember that the dot means to multiply. $\text{The sequence is given as }u_n=2^{n+2}+3^{2n+1}\\ \text{and we want to reverse engineer to show that }u_{n+1}=2u_n+7\cdot 3^{2n+1}\\ \text{Direct substitution will be all that's required.}\\ \text{Be careful when using index laws and basic algebra.}$
$\begin{align*}u_{n+1}&=2^{n+3}+3^{2n+3}\\ &= 2\cdot 2^{2n+2}+9\cdot 3^{2n+1}\\&= 2 \cdot 2^{2n+2} + 2 \cdot 3^{2n+1} + 7\cdot 3^{2n+1}\\ &= 2\left(2^{n+2}+3^{2n+1}\right)+7\cdot 3^{2n+1}\\ &= 2u_n+7\cdot 3^{2n+1} \end{align*}\\ \text{as required.}$
$\text{Observe that when }n=1\\ u_1=2^3+3^3=35=7\cdot 5\\ \text{which is divisible by 7}$
$\text{If we are to assume the proposition is true for some }n=k\\ \text{Then we have }u_k=7M\text{ for some integer }M\\ \text{We must use this truth to prove that the case }n=k+1\text{ holds true, which is easy.}$
$\text{By using our assumption, and what we proved in advance:}\\ \begin{align*}u_{k+1}&=2u_k+7\cdot 3^{2n+1}\\ &= 2(7M)+7\cdot 3^{2n+1}\\ &= 7(2M + 3^{2n+1})\end{align*}\\ \text{which is divisible by 7}\\ \text{So by the principle of mathematical induction, }u_n\text{ is divisible}\\ \text{by 7 for all integers }n\ge 1$

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Author Topic: Harder induction (Read 1122 times) Tweet Share

hanaacdr

Harder induction

RuiAce

Re: Harder induction

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