Question

If n is any positive integer, show that the integral part of $(3+\sqrt{7}{)}^n$ is an odd number.

Answer 1

suppose for $(3+\sqrt{7}{)}^n$

I = integral part

f = fractional part

$(3+\sqrt{7}{)}^n=I+f$

$0<f<1$

$3-\sqrt{7}<1$

$\therefore (3-\sqrt{7}{)}^n=f^{\prime }$ a proper fraction

adding both equation

$(3+\sqrt{7}{)}^n+(3-\sqrt{7}{)}^n=I+f+f^{\prime }$

$[C_0{3}^n.(\sqrt{7}{)}^0+C_1{3}^{n-1}.(\sqrt{7}{)}^1+.......C_n{3}^0.(\sqrt{7}{)}^n]+[C_0{3}^n.(\sqrt{7}{)}^0-C_1{3}^{n-1}.(\sqrt{7}{)}^1+.......C_n{3}^0.\left(\sqrt{7}{)}^n\right]=I+f+f^{\prime }$

$f+f^{\prime }=1$

let us assume n = even

$2.[C_0{3}^n.(\sqrt{7}{)}^0+C_1{3}^{n-2}.(\sqrt{7}{)}^2+.......C_n{3}^0.(\sqrt{7}{)}^n]=I+1$

$2.[C_0{3}^n.(\sqrt{7}{)}^0+C_1{3}^{n-2}.(\sqrt{7}{)}^2+.......C_n{3}^0.(\sqrt{7}{)}^n]=even$ integer

$\therefore I=even-1=odd$ integer