1. 如果一个正整数可以表示为两个连续奇数的平方差,那么称该正整数为“和谐数”(如 $8 = 3^{2}-1^{2}$,$16 = 5^{2}-3^{2}$,即 8,16 均为“和谐数”),在不超过 2023 的正整数中,所有的“和谐数”之和为(
D
)
A.255054
B.255064
C.250554
D.255024
答案：D 点拨:由$(2n+1)^{2}-(2n-1)^{2}=8n≤2023$,解得$n≤252\frac {7}{8}$,则在不超过 2023 的正整数中,所有的“和谐数”之和为$3^{2}-1^{2}+5^{2}-3^{2}+... +505^{2}-503^{2}=505^{2}-1^{2}=255024$.
2. (2024 春·秦都区月考)先阅读下列材料,然后解答问题：
某同学在计算 $3(4 + 1)(4^{2}+1)$ 时,把 3 写成 $(4 - 1)$ 后发现可以连续运用平方差公式计算. 即 $3(4 + 1)(4^{2}+1)= (4 - 1)(4 + 1)(4^{2}+1)= (4^{2}-1)(4^{2}+1)= 16^{2}-1$. 很受启发,后来再求 $(2 + 1)(2^{2}+1)(2^{4}+1)(2^{8}+1)…(2^{2048}+1)$ 的值时,又改造此法,将乘积式前面乘 1,且把 1 写成 $(2 - 1)$,
即 $(2 + 1)(2^{2}+1)(2^{4}+1)(2^{8}+1)…(2^{2048}+1)$
$=(2 - 1)(2 + 1)(2^{2}+1)(2^{4}+1)(2^{8}+1)…(2^{2048}+1)$
$=(2^{2}-1)(2^{2}+1)(2^{4}+1)(2^{8}+1)…(2^{2048}+1)= (2^{4}-1)(2^{4}+1)(2^{8}+1)…(2^{2048}+1)$
$=(2^{2048}-1)(2^{2048}+1)= 2^{4096}-1$.
问题：
(1) 请借鉴该同学的经验,计算：$\frac{3}{2}(1+\frac{1}{2^{2}})(1+\frac{1}{2^{4}})(1+\frac{1}{2^{8}})+\frac{1}{2^{15}}$；
(2) 借用上述的方法,再逆用平方差公式计算：
$(1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})(1-\frac{1}{4^{2}})(1-\frac{1}{5^{2}})…(1-\frac{1}{n^{2}})$.(n 为自然数,且 $n\geq2$)
答案：解:(1)原式$=2(1-\frac {1}{2})(1+\frac {1}{2})(1+\frac {1}{2^{2}})(1+\frac {1}{2^{4}})\cdot(1+\frac {1}{2^{8}})+\frac {1}{2^{15}}$$=2(1-\frac {1}{2^{2}})(1+\frac {1}{2^{2}})(1+\frac {1}{2^{4}})(1+\frac {1}{2^{8}})+\frac {1}{2^{15}}$$=2(1-\frac {1}{2^{4}})(1+\frac {1}{2^{4}})(1+\frac {1}{2^{8}})+\frac {1}{2^{15}}$$=2(1-\frac {1}{2^{8}})(1+\frac {1}{2^{8}})+\frac {1}{2^{15}}$$=2(1-\frac {1}{2^{16}})+\frac {1}{2^{15}}$$=2-\frac {1}{2^{15}}+\frac {1}{2^{15}}$$=2.$
(2)$(1-\frac {1}{2^{2}})(1-\frac {1}{3^{2}})(1-\frac {1}{4^{2}})(1-\frac {1}{5^{2}})... (1-\frac {1}{n^{2}})$$=(1+\frac {1}{2})(1-\frac {1}{2})(1+\frac {1}{3})(1-\frac {1}{3})(1+\frac {1}{4})(1-\frac {1}{4})...(1+\frac {1}{n})(1-\frac {1}{n})$$=\frac {3}{2}×\frac {1}{2}×\frac {4}{3}×\frac {2}{3}×\frac {5}{4}×\frac {3}{4}×... ×\frac {n+1}{n}×\frac {n-1}{n}$$=\frac {1}{2}×\frac {n+1}{n}$$=\frac {n+1}{2n}.$
3. 若 $x\neq y$,且 $x^{2}-4x + y = 0$,$y^{2}-4y + x = 0$,求 $x^{3}+2xy + y^{3}$ 的值.
答案：解:$\because x^{2}-4x+y=0$①,$y^{2}-4y+x=0$②,①-②,得:$x^{2}-4x+y-y^{2}+4y-x=0,$即$x^{2}-y^{2}-5(x-y)=0,$$\therefore (x+y)(x-y)-5(x-y)=0,$$\therefore (x-y)(x+y-5)=0.$$\because x≠y,\therefore x-y≠0,\therefore x+y-5=0$,即$x+y=5.$$\because x^{2}-4x+y=0,y^{2}-4y+x=0,$$\therefore x^{2}=4x-y,y^{2}=4y-x,$$\therefore x^{3}+2xy+y^{3}=x\cdot x^{2}+2xy+y\cdot y^{2}$$=x(4x-y)+2xy+y(4y-x)$$=4x^{2}-xy+2xy+4y^{2}-xy$$=4(x^{2}+y^{2})$$=4(4x-y+4y-x)$$=4(3x+3y)$$=12(x+y)$$=12×5$$=60.$
解析：
解：已知$x^{2}-4x + y = 0$①，$y^{2}-4y + x = 0$②，
① - ②得：$x^{2}-4x + y - (y^{2}-4y + x) = 0$，
即$x^{2}-y^{2}-5x + 5y = 0$，
因式分解得：$(x - y)(x + y) - 5(x - y) = 0$，
$(x - y)(x + y - 5) = 0$，
因为$x≠y$，所以$x - y≠0$，则$x + y - 5 = 0$，即$x + y = 5$。
由①得$x^{2}=4x - y$，由②得$y^{2}=4y - x$，
所以$x^{3}+2xy + y^{3}=x\cdot x^{2}+2xy + y\cdot y^{2}$
$=x(4x - y)+2xy + y(4y - x)$
$=4x^{2}-xy + 2xy + 4y^{2}-xy$
$=4x^{2}+4y^{2}$
$=4(x^{2}+y^{2})$
$=4[(4x - y)+(4y - x)]$
$=4(3x + 3y)$
$=12(x + y)$
$=12×5 = 60$。
故$x^{3}+2xy + y^{3}$的值为$60$。
