Prove this by Mathematical Induction?

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

1 Answer
George C.
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#

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