题目\LARGE题目

$\ \\\ \\\displaystyle1.计算\frac12+\frac16+\frac1{12}+\frac1{20}+\frac1{30}+\frac1{42}.$

$\ \\\displaystyle2.计算\frac12+\frac16+\frac1{12}+\frac1{20}+\cdots.$

$\ \\\displaystyle3.计算\sum_{i=1}^{100}i(i+1),即1\times2+2\times3+3\times4+\cdots+100\times101.$

$\ \\\displaystyle4.计算\sum_{i=0}^{n-1}\frac{(i+k)!}{i!},即1\times2\times3\times\cdots\times k+2\times3\times4\times\cdots\times(k+1)+\cdots+n\times(n+1)\times(n+2)\times\cdots\times(n+k-1).$

 5.计算∑i=1ni,即1+2+⋯+n.\ \\\displaystyle5.计算\sum_{i=1}^ni,即1+2+\cdots+n.

 6.计算∑i=1ni2.\ \\\displaystyle6.计算\sum_{i=1}^ni^2.

 7.计算∑i=1ni3.\ \\\displaystyle7.计算\sum_{i=1}^ni^3.

$\ \\\displaystyle8.设f(n)=\sum_{i=1}^mi^n\;(n\in\N),求f(n)的递推公式.\\\ \\\,$

$\LARGE题解$

  1.\ \\\ \\1.

$$\begin{align*} 原式&=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}+\frac{7-6}{6\times7}\\ &=\left(\frac2{1\times2}-\frac1{1\times2}\right)+\left(\frac3{2\times3}-\frac2{2\times3}\right)+\cdots+\left(\frac7{6\times7}-\frac6{6\times7}\right)\\ &=1-\frac12+\frac12-\frac13+\frac13-\frac14+\frac14-\frac15+\frac15-\frac16+\frac16-\frac17\\ &=1-\frac17\\ &=\frac67. \end{align*}$$

 2.因为\ \\2.因为

$$\begin{align*} &\frac12+\frac16+\frac1{12}+\frac1{20}+\cdots+\frac1{n(n+1)}\\ =\;&\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\left(\frac14-\frac15\right)+\cdots+\left(\frac1n-\frac1{n+1}\right)\\ =\;&1-\frac1{n+1}, \end{align*}$$

当n越来越大时,1−1n+1越来越接近1,故原式=1.\displaystyle当n越来越大时,1-\frac1{n+1}越来越接近1,故原式=1.

$\ \\\displaystyle3.i(i+1)=\frac13[i(i+1)(i+2)-(i-1)i(i+1)],$

$$\begin{align*} 原式&=\sum_{i=1}^{100}\frac13[i(i+1)(i+2)-(i-1)i(i+1)]\\ &=\frac13\left[\sum_{i=1}^{100}i(i+1)(i+2)-\sum_{i=1}^{100}(i-1)i(i+1)\right]\\ &=\frac13\left[\sum_{i=1}^{100}i(i+1)(i+2)-\sum_{i=0}^{99}i(i+1)(i+2)\right]\\ &=\frac13\left\{\left[\sum_{i=1}^{99}i(i+1)(i+2)+100\times101\times102\right]-\left[\sum_{i=1}^{99}i(i+1)(i+2)+0\right]\right\}\\ &=\frac13\times100\times101\times102\\ &=343400. \end{align*}$$

 4.\ \\4.

$$\begin{align*} \frac{(i+k)!}{i!}&=\frac1{k+1}\left[\frac{(i+k)!}{i!}\cdot(i+k+1)-\frac{(i+k)!}{i!}\cdot i\right]\\ &=\frac1{k+1}\left[\frac{(i+k+1)!}{i!}-\frac{(i+k)!}{(i-1)!}\right]\;(i>0),\\ 原式&=k!+\sum_{i=1}^{n-1}\frac1{k+1}\left[\frac{(i+k+1)!}{i!}-\frac{(i+k)!}{(i-1)!}\right]\\ &=k!+\frac1{k+1}\left[\sum_{i=1}^{n-1}\frac{(i+k+1)!}{i!}-\sum_{i=1}^{n-1}\frac{(i+k)!}{(i-1)!}\right]\\ &=k!+\frac1{k+1}\left[\sum_{i=1}^{n-1}\frac{(i+k+1)!}{i!}-\sum_{i=0}^{n-2}\frac{(i+k+1)!}{i!}\right]\\ &=k!+\frac1{k+1}\left\{\left[\sum_{i=1}^{n-2}\frac{(i+k+1)!}{i!}+\frac{(n+k)!}{(n-1)!}\right]-\left[\sum_{i=1}^{n-2}\frac{(i+k+1)!}{i!}+(k+1)!\right]\right\}\\ &=k!+\frac1{k+1}\left[\frac{(n+k)!}{(n-1)!}-(k+1)!\right]\\ &=k!+\frac{(n+k)!}{(k+1)(n-1)!}-k!\\ &=\frac{(n+k)!}{(k+1)(n-1)!}. \end{align*}$$

 5.\ \\5.

$$\begin{align*} 原式&=\frac12\left(\sum_{i=1}^ni+\sum_{i=1}^ni\right)\\ &=\frac12\left[\sum_{i=1}^ni+\sum_{i=1}^n(n-i+1)\right]\\ &=\frac12\sum_{i=1}^n[i+(n-i+1)]\\ &=\frac12\sum_{i=1}^n(n+1)\\ &=\frac{n(n+1)}2. \end{align*}$$

 6.a3−(a−1)3=a3−(a3−3a2+3a−1)=3a2−3a+1,\ \\6.a^3-(a-1)^3=a^3-(a^3-3a^2+3a-1)=3a^2-3a+1,

$$\begin{align*} 原式&=\frac13\left[\sum_{i=1}^n(3i^2-3i+1)+3\sum_{i=1}^ni-\sum_{i=1}^n1\right]\\ &=\frac13\left\{\sum_{i=1}^n[i^3-(i-1)^3]+3\cdot\frac{n(n+1)}2-n\right\}\\ &=\frac13\left[\sum_{i=1}^ni^3-\sum_{i=1}^n(i-1)^3+\frac32n^2+\frac32n-n\right]\\ &=\frac13\left(\sum_{i=1}^ni^3-\sum_{i=0}^{n-1}i^3+\frac32n^2+\frac12n\right)\\ &=\frac16[2(n^3-0)+3n^2+n]\\ &=\frac16(2n^3+3n^2+n)\\ &=\frac{n(n+1)(2n+1)}6. \end{align*}$$

 7.a4−(a−1)4=4a3−6a2+4a−1,\ \\7.a^4-(a-1)^4=4a^3-6a^2+4a-1,

$$\begin{align*} 原式&=\frac14\left[\sum_{i=1}^n(4i^3-6i^2+4i-1)+6\sum_{i=1}^ni^2-4\sum_{i=1}^ni+\sum_{i=1}^n1\right]\\ &=\frac14\left\{\sum_{i=1}^n[i^4-(i-1)^4]+6\cdot\frac{n(n+1)(2n+1)}6-4\cdot\frac{n(n+1)}2+n\right\}\\ &=\frac14\left[\sum_{i=1}^ni^4-\sum_{i=1}^n(i-1)^4+2n^3+n^2\right]\\ &=\frac14\left(\sum_{i=1}^ni^4-\sum_{i=0}^{n-1}i^4+2n^3+n^2\right)\\ &=\frac14[(n^4-0)+2n^3+n^2]\\ &=\frac14(n^4+2n^3+n^2)\\ &=\frac{n^2(n+1)^2}4. \end{align*}$$

$\ \\\displaystyle8.(a-1)^{n+1}=\sum_{i=0}^{n+1}(-1)^i\binom{n+1}ia^{n-i+1},a^{n+1}-(a-1)^{n+1}=\sum_{i=1}^{n+1}(-1)^{i+1}\binom{n+1}ia^{n-i+1}=\sum_{i=0}^n(-1)^i\binom{n+1}{i+1}a^{n-i},$

$$\begin{align*} 原式&=\frac1{\displaystyle\binom{n+1}1}\left[\sum_{i=1}^m\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}i^{n-j}-\sum_{i=1}^m\sum_{j=1}^n(-1)^j\binom{n+1}{j+1}i^{n-j}\right]\\ &=\frac1{n+1}\left\{\sum_{i=1}^m[i^{n+1}-(i-1)^{n+1}]+\sum_{i=1}^n\sum_{j=1}^m(-1)^{i+1}\binom{n+1}{i+1}j^{n-i}\right\}\\ &=\frac1{n+1}\left[m^{n+1}+\sum_{i=1}^n(-1)^{i+1}\binom{n+1}{i+1}f(n-i)\right]. \end{align*}$$

$\displaystyle记S_n=\sum_{i=1}^n(-1)^{i+1}\binom{n+1}{i+1}f(n-i),则f(n)=\frac{S_n+m^{n+1}}{n+1}.$

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