[hello]

Hi. This is a question from the Heinemann Edexcel P6 Book. Review Exercise, question 62 (page 175).

Prove that, for all positive integers, n, the following is divisible by 25:

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

I've tried the standard technique that I was taught, that is, by considering:

f(k+1) - f(k)

...and attempting to obtain a factor of 25 from the resulting expression, but was unable to do so.

Hope someone can help with this! Cheers.

[dvs]

Try f(k+1) + f(k) instead.

[Gaz031]

Prove that, for all positive integers, n, the following is divisible by 25:
7^(2n) + (2^(3n-3))(3^(n-1))

f(n)=7^(2n)+[2^3(n-1)][3^(n-1)]
f(n)=7^(2n)+[8^(n-1)][3^(n-1)]
f(n)=7^(2n)+24^(n-1).

Let the statement be true for n=k: f(k)=[7^(2k)]+[24^(k-1)]=25y
f(k+1)=7^(2k+2)+24^(k)
=(49)(7^2k)+24[24^(k-1)]
=(50)(7^2k)+25[24^(k-1)]-f(k)
=25[2(7^2k)+24^(k-1)]-f(k)
So if f(k) is divisible by 25 then so is f(k+1).

For n=1: f(1)=49+1=50=(2)(25).

Hence the statement is true for n=1, n=1+1=2, n=2+1=3.... and so on for all positive integral n.

[hello]

Thanks guys. You're quick, too!