# mathematical induction problem

• Feb 24th 2009, 03:00 PM
oblixps
mathematical induction problem
Use mathematical induction to prove the following:
a factor of (9^n - 8n - 1) is 64 for all n > or = 2

9^1 - 8 - 1 = 64a
64 = 64a

where a = any integer

9^k - 8k - 1 = 64a

prove that 9^(k+1) - 8(k+1) - 1 = 64b; where b = any integer
9(9^k) - 8k - 9

• Feb 24th 2009, 03:18 PM
arpitagarwal82
First show that the statement is true for n = 2. Which infact is.

Now assume

\$\displaystyle 9^k - 8k - 1\$ is true.

Consider

\$\displaystyle 9^ (k + 1) - 8(k + 1) - 1\$

Which is same as
\$\displaystyle 9 * 9^k - 8k - 9\$

which on simplification gives \$\displaystyle 9 (9^k - 8k - 1) - 64k\$

Now by assumption \$\displaystyle 9 (9^k - 8k - 1) \$is divisible by 64 and for k > 2 64k is divisible by 64.
So statement is true for k + 1

Hence proved.

P.S I am new to this forum and am learning how to type mathematical expressions. So ignore any typing mistake.
• Feb 24th 2009, 03:22 PM
skeeter
Quote:

Originally Posted by oblixps
Use mathematical induction to prove the following:
a factor of (9^n - 8n - 1) is 64 for all n > or = 2

9^1 - 8 - 1 = 64a
64 = 64a

where a = any integer

9^k - 8k - 1 = 64a

prove that 9^(k+1) - 8(k+1) - 1 = 64b; where b = any integer

9(9^k) - 8k - 9

\$\displaystyle 9(9^k) - 9 - 8k\$

\$\displaystyle 9(9^k - 1) - 8k\$

\$\displaystyle 9(9^k - 8k - 1) - 8k + 9(8k)\$

\$\displaystyle 9(9^k - 8k - 1) + 8(8k)\$

\$\displaystyle 9(9^k - 8k - 1) + 64k\$

\$\displaystyle 9(64a) + 64k\$

\$\displaystyle 64(9a + k)\$
• Feb 24th 2009, 03:29 PM
arpitagarwal82
Thanks skeeter. That was very clear way to explain. I went bit gross in my explanation. :)
• Feb 24th 2009, 06:06 PM
oblixps
thank you very much for the help guys! (Nod)