Split - Series question

[Sonal7]

I might as well put a proper one up, as I think Q 13 is challenging and I don't know where to go next

 

[Deleted member 4993]

I might as well put a proper one up, as I think Q 13 is challenging and I don't know where to go next

U_n = 11^(n+2) + 12^(2*n + 1)...........................Those parentheses are important!
Continue.....

 

[pka]

I might as well put a proper one up, as I think Q 13 is challenging and I don't know where to go next

You need to learn to do what the problem asks.
Is says to prove that \(\displaystyle 133\left| {\left( {{{11}^{n + 2}} + {{12}^{2n + 1}}} \right),n \geqslant 0} \right.\)
So the base case is for \(\displaystyle n=0\) or \(\displaystyle {11^{0 + 2}} + {12^{2 \cdot 0 + 1}} = 121 + 12 = 133\text{ TRUE}.\)
Now suppose that \(\displaystyle 11^{K + 2} + 12^{2 \cdot K + 1}\text{ is divisible by }133\) so prove for \(\displaystyle n=K+1.\)

 

[Steven G]

Now factor out 11 from the 1st term and factor out 12² from the 2nd term. Then be clever!