高等数学 级数的求和、函数.pdf

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1、一、某些级数的部分和（小孩，，，像下面的要证明的话，像下面的要证明的话，，，就用数学归纳法，就用数学归纳法！！！）11+2+3+L+n=n(n+)12112+22+32+L+n2=n(n+1)(2n+)16113+23+33+L+n3=n2(n+)1244444121+2+3+L+n=n(n+1)(2n+1)(3n+3n−)130555512221+2+3+L+n=n(n+)12(n+2n−)11266661431+2+3+L+n=n(n+1)(2n+1)(3n+6n−3n+)14277771224321+2+3+L+n=n(n+)13(n+6n−n−4n+)22

2、41(n+1),n为奇数n−121−2+3−L+(−)1n=n−,n为偶数2112−22+32−L+(−)1n−1n2=(−)1n−1n(n+)12122(n−1)(n+)1,n为奇数333n−1341−2+3−L+(−)1n=12−n2(n+3),n为偶数4114−24+34−L+(−)1n−1n4=(−)1n−1n(n+1)(n2+n−)122+4+6+L+2n=n(n+)121+3+5+L+2(n−)1=n2222121+3+5+L+2(n−)1=n4(n−)133333221+3+5+L+2(n−)1=n2(n−)111⋅2+2⋅3+3

3、⋅4+L+n(n+)1=n(n+1)(n+)2311⋅2⋅3+2⋅3⋅4+3⋅4⋅5+L+n(n+1)(n+)2=n(n+1)(n+2)(n+)3411⋅2⋅3⋅4+2⋅3⋅4⋅5+L+n(n+1)(n+2)(n+)3=n(n+1)(n+2)(n+3)(n+)45n1(n+k+1)!∑j(j+)1L(j+k)=j=1k+2(n−1)!n21∑j(j+)1=n(n+1)(n+2)(3n+)5j=112n21∑j(j+)1(j+)2=n(n+1)(n+2)(n+3)(2n+)3j=110n22122∑j(n−j)=n(n−)1j=14njn+12∑2j(j+)1=2(

4、n−n+)2−4j=111111n+++L+=1−=1⋅22⋅33⋅4n(n+)1n+1n+1111111+++L+=−1⋅2⋅32⋅3⋅43⋅4⋅5n(n+1)(n+)24(2n+1)(n+)21111+++L+1⋅2⋅3⋅42⋅3⋅4⋅53⋅4⋅5⋅6n(n+1)(n+2)(n+)311=−18(3n+1)(n+2)(n+)3nn11311∑=∑2=−−j=2(j+1)(j−)1j=2j−142n(2n+)1n1n∑=j=12(j−1)(2j+)12n+1n1n∑=j=13(j−2)(3j+)13n+1n111∑=−j=12(j−1)(2j+1)(2j+)31

5、22(4n+1)(2n+)3n111∑=−j=13(j−2)(3j+1)(3j+)4243(6n+1)(3n+)4n2j−1321∑=−+j=1j(j+1)(j+)24n+2(2n+1)(n+)2nj+229134∑=−−−j=1j(j+1)(j+)336n+3(2n+2)(n+)3(3n+1)(n+2)(n+)3nj−1nj221∑=−j=1(j+1)(j+)2n+22n2jn+1j42(n−4)1∑=+j=1(j+1)(j+)23(3n+)2nj+21∑=1−jnj=1j(j+2)1(n+2)1n2j+31∑=1−jnj=1j(j+3)1(n+3)1nj−1j

6、n+1(−)121(−)1∑jjj+1j+1=1+n+1n+1j=12[+(−)1][2+(−)1]32+(−)1nb(b+)1L(b+j−)11b(b+)1L(b+n)∑=−bj=1a(a+)1L(a+j−)1b−a+1a(a+)1L(a+n−)1二、乘法与因式分解公式（容易推导）2(x+a)(x+b)=x+(a+b)x+ab222(a±b)=a±2ab+b33223(a±b)=a±3ab+3ab±b22a−b=(a−b)(a+b)3322a±b=(a±b)(amab+b)nnn−1n−2n−32n−2n−1a−b=(a−b)(a+ab+

7、ab+L+ab+b)(n为正整数)nnn−1n−2n−32n−2n−1a−b=(a+b)(a−ab+ab−L+ab−b)(n为偶数)nnn−1n−2n−32n−2n−1a+b=(a+b)(a−ab+ab−L−ab+b)(n为奇数)2222(a+b+c)=a+b+c+2ab+2bc+2ca333222a+b+c−3abc=(a+b+c)(a+b+c−ab−bc−ca)三、有关三角函数、指数函数、对数函数的不等式πsinx−1−x(−∞

8、<∞,x≠