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a,Đặt \(A=1+2+2^2+2^3+...+2^{100}\)
\(2A=2+2^2+2^3+2^4+...+2^{101}\)
\(A=2^{101}-1\)
b, Đặt \(B=5+5^3+5^5+...+5^{99}\)
\(25B=5^3+5^5+5^7+...+5^{101}\)
\(24B=5^{101}-5\)
\(B=\frac{5^{101}-5}{24}\)
Đặt \(A=1+2+2^2+2^3+...+2^{100}\)
=>\(2A=2+2^2+2^3+...+2^{101}\)
=>\(2A-A=A=\text{}\text{}2+2^2+2^3+...+2^{101}-1-2-2^2-...-2^{100}=2^{101}-1\)
\(A=1+2+2^2+...+2^{100}\)
\(2A=2+2^2+2^3+...+2^{101}\)
\(2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...+2^{100}\right)\)
\(A=2^{101}-1\)
\(B=5+5^3+...+5^{99}\)
\(25B=5^3+5^5+...+5^{101}\)
\(25B-B=\left(5^3+5^5+...+5^{101}\right)-\left(5+5^3+...+5^{99}\right)\)
\(24B=5^{101}-5\)
\(B=\frac{5^{101}-5}{25}=\frac{5^{100}-1}{5}\)
\(A=1+2+2^2+....+2^{100}\)
\(\Leftrightarrow2A=2+2^2+.....+2^{100}+2^{101}\)
\(\Leftrightarrow2A-A=\left(2+2^2+....+2^{101}\right)-\left(1+2+....+2^{100}\right)\)
\(\Leftrightarrow A=2^{101}-1\)
\(B=5+5^3+.....+5^{97}+5^{99}\)
\(\Leftrightarrow5^2B=5^3+5^5+....+5^{99}+5^{101}\)
\(\Leftrightarrow25B-B=\left(5^3+5^5+....+5^{101}\right)-\left(5+5^3+...+5^{97}\right)\)
\(\Leftrightarrow24B=5^{101}-5\)
\(\Leftrightarrow B=\frac{5^{101}-5}{24}\)
(12+22+32+...+102).(1+3+5...+97+99).(36.333-108.111) [ Đặt (12+22+32+...+102).(1+3+5...+97+99) = A]
= A. (36.3.111-108.111)
= A.(108.111-108.111)
= A.0
= 0
Vậy tích trên bằng 0
(12+22+32+...+102).(1+3+5...+97+99).(36.333-108.111)
= ( 12 + 22 + 32 + ... + 102 ) . ( 1 + 3 +5 + ... + 97 + 99 ) . ( 36 . 3 . 111 - 36 . 3 . 111 )
= ( 12 + 22 + 32 + ... + 102 ) . ( 1 + 3 +5 + ... + 97 + 99 ) . 0
= 0
1. 1-2+3-4+5-6-.....+99-100
=(1-2)+(3-4)+(5-6)+...+(99-100) (50 cặp)
=(-1)+(-1)+(-1)+...+(-1) (50 số -1)
=(-1).50
=-50
2.1+3-5-7+9+11-.....-397-399
=(1+3-5-7)+(9+11-13-15)+....+(387+389-391-393)+395-397-399 (99 cặp)
=(-8)+(-8)+(-8)+...+(-8)+(-401)(có 99 có -8)
=(-8).99+(-401)
=(-792)+(-401)
=-1193
3. 1-2-3+4+5-6-7+...+96+97-98-99+100
=(1-2-3+4)+(5-6-7+8)+...+(93-94-95+96)+(97-98-99+100) (25 cặp)
=0+0+0+...+0
=0
4. A=2100-299-298-.....-22-2-1
2A=2101-2100-299-....-23-22-2
2A-A=A=2101-2100-2100+1
A=2101-2.2100+1
A=2101-2101+1
A=1
\(a,\)Đặt \(A=1+2+2^2+...+2^{99}+2^{100}\)
\(\Rightarrow2A=2+2^2+...+2^{100}+2^{101}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...2^{100}\right)\)
\(\Rightarrow A=2^{101}-1\)
\(b,\)Đặt \(B=5+5^3+5^5+...+5^{97}+5^{99}\)
\(\Rightarrow5^2B=5^3+5^5+...+5^{99}+5^{101}\)
\(\Rightarrow25B-B=\left(5^3+5^5+...+5^{99}+5^{101}\right)-\left(5+5^3+...+5^{99}\right)\)
\(\Rightarrow24B=5^{101}-5\)
\(\Rightarrow B=\frac{5^{101}-5}{24}\)
`1+3+5+...+97+99`
Số số hạng dãy trên :
`(99-1):2+1=50` (số hạng)
Tổng dãy trên :
\(\left(99+1\right)\times50:2=2500\)
Suy ra : \(\left(x+2\right)^2=2500=\left(\pm50\right)^2\\ =>x+2=\pm50\)
Vậy giá trị của `x` là : `48;-52`