Browse Questions

# Show that $9^{n+1}-8n-9$ is divisible by 64 whenever n is a positive integer.

Can you answer this question?

Writing the binomial expansion of $(1+x)^{n+1}$ we have
$(1+x)^{n+1}=C(n+1,0)+C(n+1,1)x+C(n+1,2)x^2+C(n+1,3)x^3+.......C(n+1,n+1)86{n+1}$
Putting $x=8$ we get,
$(1+8)6{n+1}=C(n+1,0)+C(n+1,1)8+C(n+1,2)(8)^2+C(n+1,3)8^3+......+C(n+1,n+1)8^{n+1}$
$[C(n+1,0)=1$ and $C(n+1,1)=n+1]$
$(9)^{n+1}=1+8n+8+8+C(n+1,2)8^2+C(n+1,3)(8)^3+C(n+1,n+1)8^{n+1}$
Which is the required expansion of $(1+x)^{n+1}$ when $x=8$
$(9)^{n+1}=1+8n+8+C(n+1,2)8^2+C(n+1,3)8^3+......+C(n+1,n+1)8^{n+1}$
$(9)^{n+1}-8n-8=C(n+1,2)8^2+C(n+1,3)8^3+......+C(n+1,n+1)8^{n+1}$
$\Rightarrow 8^2[C(n+1,2)+C(n+1,3)8+C(n+1,4)8^2+C(n+1,n+1)8^{n-1}$]
$\Rightarrow 64\times$ Some constant quantity
Hence $9^{n+1}-8n-9$ is divisible by 64 whenever n is a positive integer.
Hence proved.
answered Jun 23, 2014