题目内容
12.计算:S=$\sqrt{1+\frac{1}{{1}^{2}}+\frac{1}{{2}^{2}}}$+$\sqrt{1+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}}$+…+$\sqrt{1+\frac{1}{10{0}^{2}}+\frac{1}{10{1}^{2}}}$的值为$\frac{10200}{101}$.分析 由题意设${a}_{n}=1+\frac{1}{{n}^{2}}+\frac{1}{(n+1)^{2}}$,同分后化简凑成完全平方式并求出$\sqrt{{a}_{n}}$,再利用裂项相消法求出式子S的值.
解答 解:由题意设${a}_{n}=1+\frac{1}{{n}^{2}}+\frac{1}{(n+1)^{2}}$=1+$\frac{{(n+1)}^{2}+{n}^{2}}{{n}^{2}(n+1)^{2}}$
=1+$\frac{2{n}^{2}+2n+1}{{n}^{2}(n+1)^{2}}$=1+$\frac{2n(n+1)+1}{{n}^{2}{(n+1)}^{2}}$=1+$\frac{2}{n(n+1)}+$$\frac{1}{{n}^{2}{(n+1)}^{2}}$
=$[1+\frac{1}{n(n+1)}]^{2}$,
∴$\sqrt{{a}_{n}}$=1+$\frac{1}{n(n+1)}$=1+$\frac{1}{n}-\frac{1}{n+1}$,
∴S=$\sqrt{1+\frac{1}{{1}^{2}}+\frac{1}{{2}^{2}}}$+$\sqrt{1+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}}$+…+$\sqrt{1+\frac{1}{10{0}^{2}}+\frac{1}{10{1}^{2}}}$
=100+(1$-\frac{1}{2}$)+($\frac{1}{2}-\frac{1}{3}$)+…+($\frac{1}{100}-$$\frac{1}{101}$)
=100+1-$\frac{1}{101}$=$\frac{10{1}^{2}-1}{101}=\frac{10200}{101}$,
故答案为:$\frac{10200}{101}$.
点评 本题考查构造数列法,以及裂项相消法求数列的前n项和,考查化简、变形能力,属于中档题.
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