Question

5^n+1+4.6^n whwn divide by 20 leaves remainder 9 prove by mathematical induction

Answer 1

hello here is a solution : 

Answer 2

Answer:

It is true for all values if n

Step-by-step explanation:

Let

$P(n)=5^{n+1}+4.6^{n}-9$

We have to prove that 20 is a factor of P(n) by mathematical induction.

Let us start:

Step1:

Check P(n) is true for 1.

$P(1)=5^{1+1}+4.6^{1}-9\\P(1)=5^{2}+4.6^{1}-9\\P(1)=25+24-9\\P(1)=40=20*2$

Hence P(n) is true for n=1

Step 2:

Assume that P(n) is true for some value of n say t

That is P(t) is true or

$5^{t+1}+4.6^{t}-9=20k$

Where k is some constant…

Step3:

We have to prove that P(t+1) is also true, then P(n) is true for every value of n.

Hence

$P(t+1)= 5^{(t+1+1)}+4*6^{t+1}-9$

$P(t+1)=5^{t+1}*5+4*6^t*6-9\\=5^{t+1}*5+4*6^t*(5+1)-9\\=5^{t+1}*5+(4*6^t)*5+(4*6^t) -9\\=5(5^{t+1}+4*6^t)+(4*6^t) -9\\=5(20k+9)+ 4*6^t -9\\=100k+45-9+4*6^{t}\\=100k+36+4*6^{t}\\=100k+4(9+*6^{t})\\=100k+4(20k-5^{t+1})\\=100k+80k-4*5*5^t\\=180k-20*5^t\\=20(9k-5^t)\\=20r$

where r is some constant.

Hence

$P(t+1)$ is also true ,

Thus P(n) is true for all values of n , and we can say that

When $5^{n+1}+4.6^{n}$ is divided by 20, we get 9 as remainder.