用數學歸納法證明1 n 1 ,用數學歸納法證明1 n 1 n 1 1 n 2 1 n 1 n N ,n

時間 2021-10-30 06:16:41

1樓:匿名使用者

n=2略

n=k時有1/k+1/(k+1)+……+1/k²>1k≥2令a=1/k+1/(k+1)+……+1/k²>1則n=k+1

1/(k+1)+1/(k+2)+……+1/(k+1)²=a-1/k+1/(k²+1)+……+1/(k+1)²因為1/(k²+1)>1/(k+1)²

1/(k²+2)>1/(k+1)²

……所以a-1/k+1/(k²+1)+……+1/(k+1)²>a-1/k+1/(k+1)²+……+1/(k+1)²

=a-1/k+(2k+1)*1/(k+1)²=a+(2k²+k-k²-2k-1)/k(k+1)²=a+[(k-1/2)²-5/4]k(k+1)²k≥2所以a+[(k-1/2)²-5/4]k(k+1)²>a>1所以n=k+1

1/(k+1)+1/(k+2)+……+1/(k+1)²>1綜上,……

2樓:

n=21/2+1/3+1/4=(6+4+3)/12=13/12>1

成立設n=k時,關係成立

1/k+1/(k+1)+...+1/k²>1

n=k+1時

左邊=1/(k+1)+1/(k+2)+...+1/k²+1/(k²+1)+...+1/(k+1)²

=1/k+1/(k+1)+1/(k+2)+...+1/k²+1/(k²+1)+...+1/(k+1)²-1/k

>1+1/(k²+1)+...+1/(k+1)²-1/k

要證明:

1/(k²+1)+...+1/(k+1)²-1/k≥0

(k+1)²-k²=2k+1項,

1/(k²+1)+...+1/(k+1)²>(2k+1)[1/(k²+1)(k²+2)...(k+1)²]^(2k+1)

>(2k+1)/(k+1)²

1/(k²+1)+...+1/(k+1)²-1/k>(2k+1)/(k+1)²-1/k

=[k(2k+1)-(k+1)²]/k(k+1)²

=(2k²+k-k²-2k-1)/k(k+1)²

=(k²-k-1)/k(k+1)²

=[(k-1/2)²-5/4]/k(k+1)²

k≥2,

k-1/2≥2-1/2=3/2

(k-1)²≥6/4

(k-1)²-5/4≥6/4-5/4=1/4>0

∴1/(k²+1)+...+1/(k+1)²-1/k≥0成立。

得證.

3樓:匿名使用者

證:n=2時,1/2 +1/3 +1/4=13/12>1,不等式成立。

假設當n=k(k∈n*)時,不等式成立,即1/k +1/(k+1)+...+1/k²>1

則當n=k+1時

1/(k+1)+1/(k+2)+...+1/k²+1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1)

=[1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1) -1/k] +[1/k +1/(k+1)+...+1/k²]

>[1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1) -1/k] +1

1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1) -1/k

>1/(k+1)²+ 1/(k+1)²+...+1/(k+1)² -1/k

=(2k+1)/(k+1)² -1/k

=[k(2k+1)-(k+1)²]/[k(k+1)²]

=(k²-k-1)/[k(k+1)²]

=(k²-k-2+1)/[k(k+1)²]

=[(k+1)(k-2)+1]/[k(k+1)²]

k≥2,(k+1)(k-2)≥0,1>0,(k+1)²>0

[(k+1)(k-2)+1]/[k(k+1)²]>0

1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1) -1/k

>1/(k+1)²+ 1/(k+1)²+...+1/(k+1)² -1/k

>01/(k+1)+1/(k+2)+...+1/k²+1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1)

>[1/(k²+1)+1/(k²+2)+...+1/(k²+2k+1) -1/k] +1

>1不等式同樣成立。

k為任意正整數,因此對於任意正整數n,不等式恆成立。

即:1/n +1/(n+1)+ 1/(n+2)+...+1/n²>1

解題思路:

1、運用數學歸納法,先解得n=2時不等式成立;再設n=k,進而證明n=k+1時不等式成立。

2、在證明過程中,用到了放縮法。

用數學歸納法證明不等式:1/n+1/(n+1)+1/(n+2)+.....+1/n^2>1(n屬於正整數且n>1)

4樓:匿名使用者

很簡單。

(1)當n=2時,1/2+1/3+1/4=13/12>1

(2)假設當n=k時,原式成立,即1/k+1/(k+1)+……1/(k^2)>1

則n=k+1時,原式左側為1/(k+1)+1/(k+2)+……1/(k+1)^2

(注意:此時,上下兩式相差不大,注意比較)

因為k>2

所以1/(k^2+1)>1/(k*(k+2))

1/(k^2+2)>1/(k*(k+2))……

1/(k*(k+2))=1/(k*(k+2))

所以1/(k^2+1)+1/(k^2+2)+……1/(k+1)^2>(k+2)/(k*(k+2))=1/k

所以1/(k+1)+1/(k+2)+……1/(k+1)^2>1/k+1/(k+1)+……1/(k^2)>1

所以當n=k+1時也成立

所以1/n+1/(n+1)+1/(n+2)+.....+1/n^2>1(n屬於正整數且n>1)

5樓:匿名使用者

書上例題 我汗 最簡單的一個

6樓:古君博僪慕

數學歸納法證明:

①n=1是顯然成立。

②假設對n=k-1成立,即1/(k+1)+1/(k+2)+…+1/(k+k)>13/24,則

1/(k+1)+1/(k+2)+…+1/(k+k)

=1/k+1/(k+1)+1/(k+2)+…+1/(k+k-2)

+1/(k+k)+1/(k+k-1)-1/k

>1/k+1/(k+1)+1/(k+2)+…+1/(k+k-2)+[1/(2k)+1/(2k)-1/k]

=1/k+1/(k+1)+1/(k+2)+…+1/(k+k-2)

>13/24

綜合①②得1/(n+1)+1/(n+2)+…+1/(n+n)>13/24

****************************************==

附加告訴你兩點:

第一:n=1時,不等式變為1/2+1/2>13/24,所以是成立的!

第二:這個不等式是泰勒級數,屬於高等數學的東西。

有些加強命題,不妨試試看:

3/4>1/(n+1)+1/(n+2)+…+1/(n+n)>13/24

1/(n+1)

+1/(n+2)+...+

1/2n≤7/10-1/(4n+1)

用數學歸納法證明 1/n+1/(n+1)+1/(n+2)+...+1/2n<1(n≥3)成立 要速度

7樓:查擾龍鬆

(1)n=3時,左式=1/3+1/4+1/5+1/6=3/4+1/5<1,等式成立

(2)假設n=k(k≥3)是等式成立,即1/k+1/(k+1)+1/(k+2)+...+1/2k<1

因為k≥3時有,1/k-[1/(2k+1)+1/2(k+1)]=(3k+2)/[k(2k+1)(2k+2)]>0,

即1/k>1/(2k+1)+1/2(k+1)

則當n=k+1時,

1/(k+1)+1/(k+2)+1/(k+3)+...+1/2k+1/(2k+1)+1/2(k+1)<1/k+1/(k+1)+1/(k+2)+...+1/2k<1

等式成立

8樓:胡雪猜

a.不等式法:記sn=1/n+1/(n+1)+.......

+1/2n, sn+1 -sn =1/(2n+2)+1/(2n+1)-1/n<0,即sn+1=3),成立,即sk <1,

因為sk+1 =sk-1/n+1/(2n+1)+1/(2k+2)

即當n=k+1,也成立,

綜合以上知道命題成立,即1/n+1/(n+1)+1/(n+2)+...+1/2n<1對n≥3成立

用數學歸納法證明:1/n+1/(1+n)+1/(n+2) +......1/n^2>1(n∈n且n>1)

9樓:海潔舜甲

證明:(1)當n=2,

1/2+1/3+1/4=13/12>1成立

(2)假設當n=k時,即

1/k+1/(k+1)+...+1/k^2>1

所以當n=k+1時,有:

1/(k+1)+...+1/k^2+1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)

=1/k+1/(k+1)+...+1/k^2+[1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)-1/k]

>1+[1/(k^2+1)+1/(k^2+2)+1/(k^2+2k+1)-1/k]

>1+[(2k+1)/(k^2+2k+1)-1/k]

=1+[(2k²+k-k²-2k-1)/(k²+2k+1)k]

=1+[(k²-k-1)/(k²+2k+1)k]

因為:k²-k-1>0(當k>2時)

(k²-k-1)/(k²+2k+1)k>0

所以:1/(k+1)+...+1/k^2+1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)

>1+0

=1所以當n=k+1原式也成立

綜上,有:

1/n+1/(n+1)+1/(n+2)+…+1/n^2>1(n>1且n是整數)

10樓:睦蕾鄭雁

n=1時,左邊=1*1=1

右邊=1/6*1*2*3=1

左邊=右邊,等式成立!

假設n=k時成立

(k>1)即:

1*k+2(k-1)+3(k-2)+…+(k-1)*2+k*1=(1/6)k(k+1)(k+2)

當n=k+1時;

左邊=1*(k+1)+2(k+1-1)+3(k+1-2)+…+(k+1-1)*2+(k+1)*1

=1*k+1*1+2(k-1)+2*1+…+k*1+k+(k+1)=[1*k+2(k-1)+…+(k-1)*2+k*1]+1+2+3+…+k+(k+1)

=(1/6)k(k+1)(k+2)+1+2+3+…+k+(k+1)=(1/6)k(k+1)(k+2)+1/2*(k+1)*(k+2)=(1/6)(k+1)(k+2)(k+3)=(1/6)(k+1)[(k+1)+1][(k+1)+2]=右邊原式也成立!

綜上可知,原式為真!

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