Question

If n is positive integer, then the integral part of (7+5√2)^2n+1

Answer 1

The binomial expansion is,

$(x+y)^n=\sum_{i=0}^{n} ^nC_i x^{n-i} y^i$

            $=^nC_0 x^ny^0+^nC_1 x^{n-1}y^1+^nC_2x^{n-2}y^2+^nC_3x^{n-3}y^3+...$

            $=x^n+nx^{n-1}y+\frac{n(n-1)}{2}x^{n-2}y^2+\frac{n(n-1)(n-2)}{3}x^{n-3}y^3+...$

Thus,

$(7+5\sqrt{2})^{2n+1}=(7)^{2n+1}+(2n+1)(7)^{2n}(5\sqrt{2})+\frac{(2n+1)(2n)}{2}(7)^{2n-1}(5\sqrt{2})^2+\frac{(2n+1)(2n)(2n-1)}{3}(7)^{2n-2}(5\sqrt{2})^3+...$

Since $(7)^{x}$ is always odd for all x∈N,

Thus, $(7)^{2n+1}$ is an odd number.

Also, $(5\sqrt{2})^m$ is a non-integer number if m is odd,

While, $(5\sqrt{2})^m$ is an even positive integer if m is even.

Thus,

$(7+5\sqrt{2})^{2n+1}$

$=odd+non-integer+even+non-integer+even...$$=(odd+even+even+...)+non-integer+non-integer+non-integer+...$

$=\text{(integer part) + (non-integer part)}$

Since after adding infinite even number to an odd number results an odd number,

Therefore, the integral part of $(7+5\sqrt{2})^{2n+1}$ is odd.

Answer 2

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