求证arctan(1
求证:arctan(1/3)+arctan(1/7)+arctan(1/13)+...求证：arctan(1/3)+arctan(1/7)+arctan(1/13)+...+arctan(1/(n2+n+1))=π/4-arctan(1/(n+1))
因为tan【arctan(1/k)-arctan[1/(k+1)]】 =[1/k-1/(k+1)]/[1+(1/k)*(1/(k+1)]=1/(k^2+k+1)，k=1,2,3,……,n。 所以 arctan[1/(k^2+k+1)]=arctan(1/k)-arctan[1/(k+1)]，k=1,2,3,……,n。 于是 arctan(1/3)+arctan(1/7)+arctan(1/13)+……+arctan(1/(n^2+n+1)) =[arctan1-arctan(1/2)]+[arctan(1/2)-arctan(1/3)]+[arctan(1/3)-arctan(1/4)]+……+[arctan(1/n)-arctan(1/(n+1))] =π/4-arctan[1/(n+1)] .
简单拆分即可：
这类题型,一般就是数学归纳法。在证明之前,先说明两个基本引理。 ctan(1/2)+arctan(1/3)=π/ ctan(1/n)-arctan(1/(n+1))=arctan(1/(n2+n+1))证明很容易,就不再细说。现在开始:当n=1时,根据基本引理1,等式成立。假设n-1时等式成立,在n的情况下,等式左边便是:π/4-arctan(1/n)+arctan(1/(n2+n+1))根据基本引理2,就是π/4-arctan(1/(n+1)),也就是等式右边。证毕。
n=1时左边=arctan(1/3),右边=π/4-arctan(1/2), tan[π/4-arctan(1/2)]=(1-1/2)/(1+1/2)=1/3, 两边都是锐角，所以等式成立。 假设n=k(k∈Z*)时等式成立，即 arctan(1/3)+arctan(1/7)+arctan(1/13)+。
。。+arctan(1/(k^+k+1))=π/4-arctan[1/(k+1)], tan[π/4-arctan[1/(k+1)] =[1-1/(k+1)]/[1+1/(k+1)] =k/(k+2), (k+1)^+(k+1)+1=k^+3k+3, 那么tan{π/4-arctan[1/(k+1)]+arctan[1/[(k+1)^+(k+1)+1]]} =[k/(k+2)+1/(k^+3k+3)]/[1-k/(k+2)*1/(k^+3k+3)] =[k(k^+3k+3)+k+2]/[(k+2)(k^+3k+3)-k] =[k^3+3k^+4k+2]/[k^3+5k^+8k+6], =(k+1)/(k+3) =tan[π/4-arctan[1/(k+1+1)], 仿上，π/4-arctan[1/(k+1)]+arctan{1/[(k+1)^+(k+1)+1]}=π/4-arctan[1/(k+2)]， ∴arctan(1/3)+arctan(1/7)+arctan(1/13)+。
。。+arctan{1/[(k+1)^+(k+1)+1]} =π/4-arctan[1/(k+1)]+arctan[1/[(k+1)^+(k+1)+1]] =π/4-arctan[1/(k+2)]， 即n=k+1时等式也成立。 综上，对任意n∈N*,等式都成立。
答：证明 arctan(1/3)+arctan(1/7)+...+arctan[1/(1+n+n^2)] =arctan(n+1)-π/4 证明 arctan[1/...详情>>
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