求和:Sn=(x+1x)2+(x2+1x2)2+…+(xn+1xn)2.-数学

题目简介

求和:Sn=(x+1x)2+(x2+1x2)2+…+(xn+1xn)2.-数学

题目详情

求和:Sn=(x+
1
x
2+(x2+
1
x2
2+…+(xn+
1
xn
2
题型:解答题难度:中档来源:不详

答案

当x=±1时,
∵(xn+class="stub"1
xn
)2=4,∴Sn=4n,
当x≠±1时,
∵an=x2n+2+class="stub"1
x2n

∴Sn=(x2+x4++x2n)+2n+(class="stub"1
x2
+class="stub"1
x4
++class="stub"1
x2n
)=
x2(x2n-1)
x2-1
+
x-2(1-x-2n)
1-x-2
+2n
=
(x2n-1)(x2n+2+1)
x2n(x2-1)
+2n,
所以当x=±1时,Sn=4n;
当x≠±1时,Sn=
(x2n-1)(x2n+2+1)
x2n(x2-1)
+2n.

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