MathLabs

Problem 3

Prove that ∑k=0n(2n+12k+1)23k\sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k} is not divisible by 55 for any integer n≥0n\ge0.
Step 2 of 5: Extract the rational part
Previous stepNext step
2nS=α,(1+2)2n+1=α+β22^nS=\alpha,\quad (1+\sqrt2)^{2n+1}=\alpha+\beta\sqrt2
Detailed analysis

Expand (1+2)2n+1(1+\sqrt2)^{2n+1}. The terms with even powers of 2\sqrt2 form α=∑j=0n(2n+12j)(2)2j=2nS\alpha=\sum_{j=0}^{n}\binom{2n+1}{2j}(\sqrt2)^{2j}=2^nS in F5\mathbb F_5, while the odd terms form β2\beta\sqrt2 for some β∈F5\beta\in\mathbb F_5.