x2+px+q=0x2+2⋅p2x+q=0(x+p2)2−p24+q=0(x+p2)2=p24−q(x+p2)2=p2−4q4x+p2=±p2−4q2x=−p±p2−4q2 x^2+px+q=0\\ x^2+2\cdot \frac{p}{2}x+q=0\\ (x+\frac{p}{2})^2-\frac{p^2}{4}+q=0\\ (x+\frac{p}{2})^2=\frac{p^2}{4}-q\\ (x+\frac{p}{2})^2=\frac{p^2-4q}{4}\\ x+\frac{p}{2}=\frac{\pm\sqrt{p^2-4q}}{2}\\
x=\frac{-p\pm\sqrt{p^2-4q}}{2}\\ x2+px+q=0x2+2⋅2p x+q=0(x+2p )2−4p2 +q=0(x+2p )2=4p2 −q(x+2p )2=4p2−4q x+2p =2±p2−4q x=2−p±p2−4q
S=a1+a2+a3+a4+⋯+an(aka1=qk)S=a1+qa1+q2a1+⋯+qn−1a1qS=qa1+q2a1+⋯+qna1(1−q)S=a1−qna1S=a1−qna11−qS=(1−qn)a11−q S=a_1+a_2+a_3+a_4+\cdots+a_n(\frac{a_k}{a_1}=q^k)\\ S=a_1+qa_1+q^2a_1+\cdots+q^{n-1}a_1\\ qS=qa_1+q^2a_1+\cdots+q^na_1\\ (1-q)S=a_1-q^na_1\\ S=\frac{a_1-q^na_1}{1-q}\\
S=\frac{(1-q^n)a_1}{1-q}\\ S=a1 +a2 +a3 +a4 +⋯+an (a1 ak =qk)S=a1 +qa1 +q2a1 +⋯+qn−1a1 qS=qa1 +q2a1 +⋯+qna1 (1−q)S=a1 −qna1 S=1−qa1 −qna1 S=1−q(1−qn)a1
最后谁能证明下面错
S1=1−1+1−1+1⋯1−S1=1−(1−1+1⋯ )=1−1+1−1+1⋯=S1S1=12S2=1−2+3−4+5−6⋯2S2=(1−2+3⋯ )+(0+1−2⋯ )=1−1+1−1+1⋯=S1S2=14S3=1+2+3+4+⋯S3−S2=0+4+0+8+⋯=4S3−S2=3S3S3=−112 S_1=1-1+1-1+1\cdots\\ 1-S_1=1-(1-1+1\cdots)=1-1+1-1+1\cdots=S_1\\ S_1=\frac{1}{2}\\ S_2=1-2+3-4+5-6\cdots\\
2S_2=(1-2+3\cdots)+(0+1-2\cdots)=1-1+1-1+1\cdots=S_1\\ S_2=\frac{1}{4}\\ S_3=1+2+3+4+\cdots\\ S_3-S_2=0+4+0+8+\cdots=4S_3\\ -S_2=3S_3\\ S_3=-\frac{1}{12} S1 =1−1+1−1+1⋯1−S1 =1−(1−1+1⋯)=1−1+1−1+1⋯=S1 S1 =21 S2 =1−2+3−4+5−6⋯2S2 =(1−2+3⋯)+(0+1−2⋯)=1−1+1−1+1⋯=S1 S2 =41 S3 =1+2+3+4+⋯S3 −S2 =0+4+0+8+⋯=4S3
−S2 =3S3 S3 =−121
