Prove this by Mathematical Induction?

  1. 2*7^n+3*5^n-5 is divisible by 24 ninNN

1 Answer
May 3, 2018

See explanation...

Explanation:

Let P(n) be the proposition:

2*7^n+3*5^n-5 is divisible by 24

Base case

Putting n=0, we find:

2*7^color(blue)(0)+3*5^color(blue)(0)-5 = 2+3-5 = 0

which is divisible by 24

So P(0) is true.

Induction step

Suppose P(n) for some n.

That is:

2*7^n+3*5^n-5 = 24k

for some integer k

Then:

2*7^(n+1)+3*5^(n+1)-5

= 7 * (2*7^n) + 5 * (3 * 5^n) - 5

= 7 * (2*7^n) + 7 * (3 * 5^n) - 7*5 + 35 - 2 * (3 * 5^n) - 5

= 7 * (2*7^n + 3*5^n-5) + 30 - 2 * (3 * 5^n)

= 7 * 24k - 2*(3*5^n-15)

= 7 * 24k - 2*3*5(5^(n-1)-1)

Note that:

(x+1)^n = sum_(k=0)^n ((n),(k)) x^(n-k) = x sum_(k=0)^(n-1) ((n),(k)) x^(n-k) + 1

Hence:

5^(n-1)-1 = (4+1)^(n-1)-1 = 4m+1-1 = 4m

for some integer m

So:

2 * 3 * 5*(5^(n-1)-1) = 2 * 3 * 5 * 4m = 24(5m)

for some integer m

Thus: 7 * 24k - 2*3*5(5^(n-1)-1) is divisible by 24 and P(n+1) holds.

Conclusion

Having shown P(0) and P(n) => P(n+1) we can conclude that P(n) for all n >= 0, i.e. for all n in NN