2 years ago

Half of j minus 5 is the sum of k and 13

Mathematics

1 answer:

barxatty [35]2 years ago

4 0

.5j - 5 = k + 13. Is that what you want?

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Answer with step-by-step explanation:

We are given that the recurrence relation

$f_n=5f_{n-4}+3f_{n-5}$

for n=5,6,7,..

Initial condition

$f_0=0,f_1=1,f_2=1,f_3=2,f_4=3$

We have to show that Fibonacci numbers satisfies the recurrence relation.

The recurrence relation of Fibonacci numbers

$f_n=f_{n-1}+f_{n-2}$ , $f_0=0,f_1=1$

Apply this

$f_n=(f_{n-2}+f_{n-3})+f_{n-2}=2f_{n-2}+f_{n-3}$

$f_n=2(f_{n-3}+f_{n-4})+f_{n-3}=3f_{n-3}+2f_{n-4}$

$f_n=3(f_{n-4}+f_{n-5})+2f_{n-4}=5f_{n-4}+3f_{n-5}$

Substitute n=2

$f_2=f_1+f_0=1+0=1$

$f_3=f_2+f_1=1+1=2$

$f_4=f_3+f_2=2+1=3$

Hence, the Fibonacci numbers satisfied the given recurrence relation .

Now, we have to show that $f_{5n}$ is divisible by 5 for n=1,2,3,..

Now replace n by 5n

$f_{5n}=5f_{5n-4}+3f_{5n-5}$

Apply induction

Substitute n=1

$f_5=5f_1+3f_0=5+0=5$

It is true for n=1

Suppose it is true for n=k

$f_{5k}=5f_{5k-4}+3f_{5k-5}$ is divisible 5

Let $f_{5k}=5q$

Now, we shall prove that for n=k+1 is true

$f_{5k+5}=5f_{5k+5-4}+3f_{5k+5-5}=5f_{5k+1}+3f_{5k}=5f_{5k+1}+3(5q)$

$f_{5k+5}=5(f_{5k+1}+3q)$

It is multiple of 5 .Therefore, it is divisible by 5.

It is true for n=k+1

Hence, the $f_{5n}$ is divisible by 5 for n=1,2,3,..

8 0

2 years ago

Half of j minus 5 is the sum of k and 13​

Half of j minus 5 is the sum of k and 13