Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{7}{4}+\frac{17}{9}+\frac{31}{16}+....+\frac{4999}{2500}\)
\(A=\frac{3+4}{2^2}+\frac{8+9}{3^2}+\frac{15+16}{4^2}+....+\frac{2499+2500}{50^2}\)
\(A=\frac{\left(4-1\right)+4}{2^2}+\frac{\left(9-1\right)+9}{3^2}+\frac{\left(16-1\right)+16}{4^2}+...+\frac{\left(2500-1\right)+2500}{50^2}\)
\(A=\frac{4+4-1}{2^2} +\frac{9+9-1}{3^2}+\frac{16+16-1}{4^2}+...+\frac{2500+2500-1}{50^2}\)
\(A=\frac{2^2+2^2-1}{2^2}+\frac{3^2+3^2-1}{3^2}+\frac{4^2+4^2-1}{4^2}+...+\frac{50^2+50^2-1}{50^2}\)
\(A=\frac{2\times2^2-1}{2^2}+\frac{2\times3^2-1}{3^2}+\frac{2\times4^2-1}{4^2}+...+\frac{2\times50^2-1}{50^2}\)
\(A=\left(\frac{2\times2^2}{2^2}-\frac{1}{2^2}\right)+\left(\frac{2\times3^2}{3^2}-\frac{1}{3^2}\right)+\left(\frac{2\times4^2}{4^2}-\frac{1}{4^2}\right)+...+\left(\frac{2\times50^2}{50^2}-\frac{1}{50^2}\right)\)
\(A=\left(\frac{2\times2^2}{2^2}+\frac{2\times3^2}{3^2}+\frac{2\times4^2}{4^2}+...+\frac{2\times50^2}{50^2}\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
\(A=\left(2+2+2+...+2\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
49 số 2
\(A=98-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
\(>98-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{49\times50}\right)\)
\(=98-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=98-\left(1-\frac{1}{50}\right)\)
\(=98-1+\frac{1}{50}\)
\(=97+\frac{1}{50}>97\)
\(\Rightarrow A>97\left(ĐPCM\right)\)
Nhớ tích cho mk nha thank for watching
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
a, Số số hạng dãy S là: (2005-1):4+1= 505 số hạng
Tổng dãy S là: (2005+1).505:2= 506515
b, 3+33+35+37+..+331
= (3+33)+34(3+33)+...+329(3+33)
= 30+34.30+...+329.30
= 30(1+34+...+329) chia hết cho 30
`Answer:`
\(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}+\frac{1}{32}\)
a) Ta thấy:
\(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}\)
\(\frac{1}{9}+...+\frac{1}{16}>8.\frac{1}{16}=\frac{1}{2}\)
\(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}>16.\frac{1}{32}=\frac{1}{2}\)
\(\Rightarrow S>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}\)
b) Ta thấy:
\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3.\frac{1}{3}\)
\(\frac{1}{6}+...+\frac{1}{11}< 6.\frac{1}{6}\)
\(\frac{1}{12}+...+\frac{1}{23}< 12.\frac{1}{12}\)
\(\frac{1}{24}+...+\frac{1}{32}< 9.\frac{1}{24}\)
\(\Rightarrow S< \frac{1}{2}+1+1+1+\frac{9}{24}=\frac{31}{8}< \frac{9}{2}\)
k)(-37)+14+26+37
=(-37)+37+14+26
=0+14+26
=40
l)15+23+(-25)+(-23)
=15+(-25)+23+(-23)
=-10+0
=-10
n)25+37-48-25-37
=25-25+37-37-48
=0+0-48
=-48
o)24.(16-5)-16.(24-5)
=24.16-24.5-16.24-16.5
=0-24.5-16.5
=0-(24-16).5
=0-8.5
=0-40
=-40
p)31.(-18)+31.(-81)-31
=31.(-18)+31.(-81)+31.(-1)
=31.[-18+(-81)+(-1)]
=31.(-100)
=-3100
q)-48+48.(-78)+48.(-21)
=48.(-1)+48.(-78)+48.(-21)
=48.[(-1)+(-78)+(-21)]
=48.(-100)
=-4800
r)(7.3-3):(-6)
=(7.3-1.3):(-6)
=(6.3):(-6)
=-3
s)-3.7-4.(-5)+1
=-21-(-20)+1
=-20-(-20)
=0
t(6.8-10:5)+3.(-7)
=(48-2)+(-21)
=46+(-21)
=25
Viết mỏi cả tay rồi.tick đi nhé
a) Xét:
5121 có chữ số tận cùng là 5. Đặt 5121 = \(\overline{A5}\)
3515 có chữ số tận cùng là 5. Đặt 3515 = \(\overline{B5}\)
Do đó \(5^{121}-35^{15}=\overline{A5}-\overline{B5}=\overline{C0}⋮10\left(đpcm\right)\)
b) Ta có:
\(\left(13-12\right)^{2015}=1^{2015}=1\)
\(5^{17}.5^{14}:5^{31}=5^0=1\)
Vậy \(\left(13-12\right)^{2015}=5^{17}.5^{14}:5^{31}\)
c) \(9+5x=4^7:4^3-3^4\)
\(\Leftrightarrow9+5x=4^4-3^4\)
\(\Leftrightarrow9+5x=256-81\)
\(\Leftrightarrow9+5x=175\)
\(\Leftrightarrow5x=175-9=166\)
\(\Rightarrow x=166:5=33\dfrac{1}{5}\)
Lời giải:
$S=(2-\frac{1}{4})+(2-\frac{1}{9})+(2-\frac{1}{16})+...+(2-\frac{1}{400})$
$=(2+2+2+....+2)-(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{400})$
$=2.19-(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{20^2})$
$> 38-(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20})$
$=38-(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20})$
$=38-(1-\frac{1}{20})=37+\frac{1}{20}> 37$
Ta có đpcm.