Mathematical Induction

**yobacul** · February 28th 2010, 09:50 AM

Prove by mathematical induction that 7^(2n) - 48n - 1 is a multiple of 2304

I am ok with the first and second steps, but I am confused with the third i.e. when substituting n = k + 1.

Any help please?
Thanks a lot

**Jhevon** · February 28th 2010, 10:18 AM

Originally Posted by yobacul

Prove by mathematical induction that 7^(2n) - 48n - 1 is a multiple of 2304

I am ok with the first and second steps, but I am confused with the third i.e. when substituting n = k + 1.

Any help please?
Thanks a lot

Hint:

Note that $7^{2(k + 1)} - 48(k + 1) - 1 = 7^{2k + 2} - 48k - 48 - 1 = 7^{2k + 2} ~\underbrace{{\color{red} - 7^2 \cdot 48k + 7^2 \cdot 48k}}_{\text{this is zero}}~ - 48k - 7^2$

**Archie Meade** · February 28th 2010, 10:51 AM

Originally Posted by yobacul

Prove by mathematical induction that 7^(2n) - 48n - 1 is a multiple of 2304

I am ok with the first and second steps, but I am confused with the third i.e. when substituting n = k + 1.

Any help please?
Thanks a lot

hi yobacul,

F(n)

$7^{2n}-48n-1$

is a multiple 2304 ?

F(k+1)

$7^{2(k+1)}-48(k+1)-1$

is a multiple of 2304 if the "k"th term is ?

Does the hypothesis F(n) cause this to be true ?

$7^{2k+2}-48k-48-1=7^2\left(7^{2k}\right)-48k-49$

$=7^2\left(7^{2k}\right)-48k-7^2=7^2\left(7^{2k}-1\right)-48k$

Now express -48k as a multiple of $7^2$

To do this we need to subtract another 48(48k)
{and therefore also add that amount}

$7^2\left(7^{2k}-1\right)-(48k)-48(48k)+48(48k)$

$=7^2\left(7^{2k}-1\right)-49(48k)+48(48k)$

$=7^2\left(7^{2k}-48k-1\right)+48^2k$

The final term is $48^2k=2304k$

therefore, the term-by-term link is established.

Hence, if F(n) is valid, F(n+1) also is

Thread: Mathematical Induction

LinkBack

Thread Tools

Display