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Lời giải:
Ta có:
\(A=1^2-2^2+3^2-4^2+....+99^2-100^2+101^2\)
\(\Leftrightarrow A=(1^2-2^2)+(3^2-4^2)+....+(99^2-100^2)+101^2\)
\(\Leftrightarrow A=(-1)(1+2)+(-1)(3+4)+....+(-1)(99+100)+101^2\)
\(\Leftrightarrow A=-(1+2+.....+99+100)+101^2\)
\(\Leftrightarrow A=-\frac{100(100+1)}{2}+101^2=101^2-50.101=101.51=5151\)
Vậy \(A=5151\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1^2-2^2+3^2-4^2+...-100^2+101^2\)
\(\left(1-2\right).\left(1+2\right)+\left(3-4\right)\left(3+4\right)\)\(+...+\left(99-100\right).\left(99+100\right)+101^2\)
\(-3-7-11-...-199+101^2\)
\(101^2-\left(3+7+11+...+199\right)\)
Ta de thay :(3+7+11+ . . .+199) la 1 cap so cong co d=4 ,n=50
\(101^2-\left(199+3\right)\cdot50:2\)
\(=5151\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1^2-2^2+3^2-4^2+.................+99^2-100^2+101^2\)
\(=\left(-3\right)+\left(-7\right)+\left(-11\right)+........+\left(-199\right)+10201\)
\(=\frac{50.\left[\left(-199\right)+\left(-3\right)\right]}{2}+10201\)
\(=\left(-5050\right)+10201\)
\(=5151\)
\(1^2-2^2+3^2-4^2+...+99^2-100^2+101^2\)
\(=\left(-3\right)+\left(-7\right)+\left(-11\right)+...+-199+101^2\)
\(=\frac{50\left(-199+\left(-3\right)\right)}{2}+10201\)
\(=-5050+10201\)
\(=5151\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A\)= 12 - 22 + 32 - 42 + ... + 992 - 1002 + 1012
\(\Leftrightarrow A\)= \(\left(1.1-2.2\right)\) \(+\)\(\left(3.3-4.4\right)\)\(+\)\(\left(5.5-6.6\right)\)\(+\)\(...\)\(+\)\(\left(99.99-100.100\right)\)\(+\)\(101.101\)
\(\Leftrightarrow A\)= \(\left(-3\right)\)\(+\)\(\left(-7\right)\)\(+\)\(\left(-11\right)\)\(+\)\(...\)\(+\)\(\left(-199\right)\)\(+\)\(10201\).Tìm số hạng của tổng.Mình tìm được 50
\(\Leftrightarrow\)\(\left(-5050\right)\)+\(10201\)=\(5151\)
chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
1, a)
Ta có:
\(x^2+2x+1=\left(x+1\right)^2\)
Thay x=99 vào ta có:
\(\left(99+1\right)^2=100^2=10000\)
b) Ta có:
\(x^3-3x^2+3x-1=\left(x-1\right)^3\)
Thay x=101 vào ta có:
\(\left(101-1\right)^3=100^3=1000000\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : $1.3+2.4+3.5+...+99.101+100.102$
$=(2-1)(2+1)+(3-1)(3+1)+(4-1)(4+1)+...+(100-1)(100+1)+(101-1)(101+1)$
$=2^2-1+3^2-1+4^2-1+...+100^2-1+101^2-1$
$=(2^2+3^2+4^2+...+100^2+101^2)-100$
b) $1.100+2.99+3.98+...+99.2+100.1$
$=1.100+2.(100-1)+3.(100-2)+...+99.(100-98)+100.(100-99)$
$=100(1+2+3+...+99+100)-(1.2+2.3+...+99.100)$
$=100.\dfrac{101.100}{2}-\dfrac{99.100.101}{3}=171700$
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=1+2+2^2+...+2^{100}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{100}+2^{101}\)
\(\Rightarrow2A-A=-1+2^{101}\)
\(\Rightarrow A=2^{101}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B1:
\(a.301^2=\left(300+1\right)^2=300^2+2.300.1+1^2\\ =90000+600+1=90601\\ b.88^2+2.88.12+12^2=\left(88+12\right)^2=100^2=10000\\ c.99.100=100^2-100=10000-100=9900\\ d,153^2+94.153+47^2=153^2+2.153.47+47^2=\left(153+47\right)^2=200^2=40000\)
B2:
\(A=x^2-20x+101\\ =x^2-2.x.10+10^2+1\\ =\left(x-10\right)^2+1\ge1\forall x\in R\left(Vì:\left(x-10\right)^2\ge0\forall x\in R\right)\\ \Rightarrow min_A=1\Leftrightarrow x-10=0\Leftrightarrow x=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a:
Đặt A=x+x^2+x^3+...+x^99+x^100
Khi x=-1 thì A=(-1)+(-1)^2+(-1)^3+...+(-1)^100
=(-1+1)+(-1+1)+...+(-1+1)
=0
b: Đặt B=x^2+x^4+...+x^100
Khi x=-1 thì B=(-1)^2+(-1)^4+...+(-1)^100
=1+1+...+1
=50
\(A=\left(1^2-2^2\right)+\left(3^2-4^2\right)+...+\left(99^2-100^2\right)+101^2\)\(=-\left(3+7+...+199\right)+101^2=-\frac{\left(3+199\right).50}{2}+101^2=5151\)