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If m is a positive integer, then [(sqrt(3)+1)^(2m)]+1, where [x] denotes greatest integer lex, must be divisible by |
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Answer» `2^(m)` `THEREFORE (I+f)+f'(sqrt(3)+1)^(2m)+(sqrt(3)-1)^(2m)impliesI+(f+f')=(4+2sqrt(3))^(m)+(4-2sqrt(3))^(m)` `impliesI+(f+f')=2^(m)[(2+sqrt(3))^(m)+(2-sqrt(3))^(m)]=I+(f+f')=2^(m)xx2[.^(m)C_(0)2^(m)+.^(m)C_(2)2^(m-2).3+.....]` `impliesI+(f+f')=2^(m+1)lambda" where " lambda in N implies f+f'in N implies f+f'=1` `therefore I+1=2^(m+1)lambdaimplies[(sqrt(3)+1)^(2m)]+1=2^(m+1)lambda` |
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