13 倒序相加与分组求和法 (基础)

说说

$\{\begin{array}{l}\mathrm{反}\mathrm{序}\mathrm{相}\mathrm{加}\mathrm{法}(\mathrm{例}\mathrm{如}\mathrm{高}\mathrm{斯}\mathrm{求}\mathrm{和})\\ \mathrm{分}\mathrm{组}\mathrm{求}\mathrm{和}(\mathrm{比}\mathrm{如}(a_n+b+2^n))\end{array}$

倒序相加

$S=si{n}^2{1}^{\circ }+si{n}^2{2}^{\circ }+...+si{n}^2{89}^{\circ }$

分组求和

$(x+\frac{1}{x}{)}^2+(x^2+\frac{1}{x^2}{)}^2+(x^3+\frac{1}{x^3}{)}^2+...+(x^n+\frac{1}{x^n}{)}^2$

$(x^{2\cdot 1}+\frac{1}{x^{2\cdot 1}+2})+(x^{2\cdot 2}+\frac{1}{x^{2\cdot 2}+2})+(x^{3\cdot 2}+\frac{1}{x^{3\cdot 2}+2})+...$

$=(x^2+x^4+x^6+...+x^{2n})+(\frac{1}{x^2}+\frac{1}{x^4}+...+\frac{1}{x^{2n}})+2n$

当 $x^2\ne 1$ 时，

$\mathrm{式}\mathrm{子}=\frac{x^2[1-(x^2)n]}{1-x^2}+\frac{\frac{1}{x^2}[1-(\frac{1}{x^2}{)}^n]}{1-\frac{1}{x^2}}+2n$

$=\frac{(x^{2n}-1)(x^{2n+2}+1)}{(x^2-1)x^{2n}}+2n$

当 $x^2=1$ 时，

$\mathrm{式}\mathrm{子}=4n$