一道数学找规律题,设S1=1加1/（1的平方）加1/（2的平方）,S2=1加1/（2的平方）加1/（3的平方）…Sn=1加1/(n的平方）加1/（n加1）的平方,设S=根号下S1加根号下S2加…根Sn.则S=多少?（答案上是（n平方加2n)/（n加1）,不明白的是前面加的那些根S1S2哪去了?
2019-04-13
一道数学找规律题,
设S1=1加1/（1的平方）加1/（2的平方）,S2=1加1/（2的平方）加1/（3的平方）…Sn=1加1/(n的平方）加1/（n加1）的平方,设S=根号下S1加根号下S2加…根Sn.则S=多少?（答案上是（n平方加2n)/（n加1）,不明白的是前面加的那些根S1S2哪去了?
优质解答
S1=1+1/1²+1/2²,
S2=1+1/2²+1/3²
…
Sn=1+1/n²+1/(n+1)²
将Sn右边通分
Sn=[n²(n+1)²+(n+1)²+n²]/[n²(n+1)²]
=[n²(n+1)²+2n(n+1)+1]/[n²(n+1)²]
=[n(n+1)+1]²/[n²(n+1)²]
∴√Sn=[n(n+1)+1]/[n(n+1)]
=1+1/[n(n+1)]
=1+1/n-1/(n+1)
∴ S=√S1+√S2+…+√Sn.
=(1+1/1-1/2)+(1+1/2-1/3)+(1+1/3-1/4)+.+[1+1/n-1/(n+1)]
=n+[1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)]
=n+1-1/(n+1)=[(n+1)²-1]/(n+1)=(n²+2n)/(n+1)
S1=1+1/1²+1/2²,
S2=1+1/2²+1/3²
…
Sn=1+1/n²+1/(n+1)²
将Sn右边通分
Sn=[n²(n+1)²+(n+1)²+n²]/[n²(n+1)²]
=[n²(n+1)²+2n(n+1)+1]/[n²(n+1)²]
=[n(n+1)+1]²/[n²(n+1)²]
∴√Sn=[n(n+1)+1]/[n(n+1)]
=1+1/[n(n+1)]
=1+1/n-1/(n+1)
∴ S=√S1+√S2+…+√Sn.
=(1+1/1-1/2)+(1+1/2-1/3)+(1+1/3-1/4)+.+[1+1/n-1/(n+1)]
=n+[1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)]
=n+1-1/(n+1)=[(n+1)²-1]/(n+1)=(n²+2n)/(n+1)