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Prove by induction that the integer next greater than `(3+sqrt(5))^n` is divisible by `2^n` for all `n in N`. |
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Answer» Let `alpha=3+sqrt(5) and beta =3-sqrt(5)` `therefore 0 lt beta^(n)lt 1,forall n in N` `rArr alpha +beta=6,alpha beta=4` .....(i) Then , `alpha and beta` are the roots of `x^2-6x+4=0` `rArr alpha^2=6alpha-4` `beta^2=6beta-4` ...(ii) `therefore alpha^2+beta^2=6(alpha+beta)-8=28` ......(iii) `therefore alpha^n+beta^n=(3+sqrt(5))^n+(3-sqrt(5))^n` `=2[3^n+.^(n)C_(2)3^(n-2).5+.^(n)C_(4)3^(n-4).5^(2)+........]` Even integer . As , `0lt beta^(n) lt 1,alpha^(n)+beta^(n)` is the even integer next greater than `alpha^(n)`. Step II For `n=1`, `alpha+beta=6` divisible by `2^1` and `n=2, alpha^2+beta^2=28` divisible by `2^2` which is true for `n=1,2`. Step II Assume it is true for `n=k` . i.e., `alpha^k+beta^k` is divisible by `2^k`. Step III For `n=k+1`. the integer next greater than `alpha^(k+1) is alpha^(k+1)+beta^(k+1)` `= alpha^2.alpha^(k-1)+beta^2.beta^(k-1)` `=(6alpha-4).alpha^(k-1)+(6beta-4).beta^(k-1)` `=6(alpha^k+beta^k)-4(alpha^(k-1)+beta_(k-1))` `=3` (divisible by `2^(k+1))-` (divisible by `2^k+1`) =Divisible by `2^k+1`. This shows that the result is true for `n=k+1`. Hence , the integer next greater than `alpha^(k+1)` is divisible by `2^(k+1)`. |
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