1/1.4 + 1/4.7 + 1/7.11 + ... + 1/n(n + 3) = 89/270
tìm n :
\(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{n\left(n+3\right)}=\frac{89}{270}\)
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\(\frac{1}{1.3}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n\left(n+3\right)}=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\right)=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}=\frac{2018}{6057}.3\)
\(\Rightarrow1-\frac{1}{n+3}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=1-\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=\frac{1}{2019}\)
\(\Rightarrow n+3=2019\)
\(\Rightarrow n=2016\)
Vậy n = 2016
=>S= 1- 1/4 + 1/4 -1/7 + 1/7 - 1/10 +...+ 1/n - 1/(n+3)
=>S= 1- 1/(n+3)
=>S + 1/(n+3) = 1
=>S<1
Ta có:
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{n+3}\)
\(\Leftrightarrow S=\frac{n+3}{n+3}-\frac{1}{n+3}=\frac{n+3-1}{n+3}=\frac{n+2}{n+3}\)
\(\Rightarrow\frac{n+2}{n+3}< 1\Rightarrow S< 1\)
c)1*(1/2-1/3+1/3-1/4+.....+1/91-1/94)
1/2-1/94 ban tu tinh nhe
d)1*(1/1-1/4+1/4-1/7+......+1/91-1/94)
1-1/94 ban tu tinh nhe
tk nha
a) \(\frac{1}{n}-\frac{1}{n+1}\left(n\inℕ^∗\right)\)
\(\Leftrightarrow\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}\Leftrightarrow\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
b) \(\frac{1}{n}-\frac{1}{n+3}\left(n\inℕ^∗\right)\)
\(\Leftrightarrow\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}=\frac{n+3-n}{n\left(n+3\right)}=\frac{3}{n\left(n+3\right)}\)
c,d dễ bn tách ra rồi trừ đi
\(\Rightarrow\) S < 1 ( đpcm )
=> S = ( 1 -\(\frac{1}{4}\)) + ( \(\frac{1}{4}\)- \(\frac{1}{7}\)) +(\(\frac{1}{7}\)- \(\frac{1}{10}\)) +.....+ (\(\frac{1}{n}\)- \(\frac{1}{n+3}\))
=> S = 1 - \(\frac{1}{4}\)+\(\frac{1}{4}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{10}\)+......+ \(\frac{1}{n}\)- \(\frac{1}{n+3}\)
=> S = 1 - \(\frac{1}{n+3}\)
vậy S = 1- \(\frac{1}{n+3}\)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{n\left(n+3\right)}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{n}-\frac{1}{n+3}\)
\(=1-\frac{1}{n+3}\)
Ta có :
\(\frac{1}{n+3}>0\)
\(\Leftrightarrow-\frac{1}{n+3}< 0\)
\(\Leftrightarrow1-\frac{1}{n+3}< 1\)
\(\Leftrightarrow S< 1\left(đpcm\right)\)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(S=1-\frac{1}{n+3}\)
\(S=\frac{n+2}{n+3}\)
Vi \(n\inℕ^∗\)nên \(n+2< n+3\)
DO đó\(\frac{n+2}{n+3}< 1\)
Vậy S <1
Ta có : \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n\left(n+3\right)}=\frac{89}{270}\)
\(\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{n\left(n+3\right)}=\frac{267}{270}\)
\(\Rightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{n}-\frac{1}{n+3}=\frac{267}{270}\)
\(\Rightarrow1-\frac{1}{n+3}=\frac{267}{270}\)
=> \(\frac{1}{n+3}=\frac{1}{90}\)
=> n + 3 = 90
=> n = 87
Nhân cả 2 vế với 3 ta được:
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{n\left(n+3\right)}=\frac{89}{90}.\)
Vậy tử số của các phân số trên đã bằng hiệu của 2 thừa số ở mẫu số.(Ngoại trừ P/S\(\frac{89}{90}.\))
=> ta được:
\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{n}-\frac{1}{n+3}=\frac{89}{90}.\)
Rút gọn hết ta được :
\(1-\frac{1}{n+3}=\frac{89}{90}\)
\(\frac{1}{n+3}=1-\frac{89}{90}\)
\(\frac{1}{n+3}=\frac{1}{90}.\)
Vì 1=1 => n+3=90
n = 90-3
n=87
Vậy n=87.
Đ/S:87