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Đặt A=( 20^2+18^2+...+2^2)-(19^2+17^2+...+1^2)
= (20^2-19^2)+ (18^2-17^2)+...+(2^2-1)
= (20-19)(20+19)+(18-17)(18+17)+...+(2-1)(2+1)
= 20+19+18+...+2+1
=20.21=...
Tương tự câu b
a) \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)-2^{16}\)
\(=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)-2^{16}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)-2^{16}\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)-2^{16}\)
\(=\left(2^8-1\right)\left(2^8+1\right)-2^{16}\)
\(=2^{16}-1-2^{16}\)
\(=-1\)
A - B = (502+482+462+.....+42+22) - (492+472+452+.....+32+12)
= 502 + 482 + 462 +... + 42+ 22 - 492 - 472 - .... - 32 - 12
= (502 - 492) + (482 - 472) + ... + (42 - 32) + (22 - 12)
= (50+49) (50 - 49) + (48 - 47) (48+47)+....+(4-3)(4+3) + (2-1)(2+1)
= 50 + 49 + 48 + 47 + 46 + 45+...+4+3+2+1
= [(50 - 1) : 1 + 1] * (50+1) : 2 = 1275
vậy A - B = 1275
\(B=\left(50^2+48^2+46^2+...+4^2+2^2\right)-\left(49^2+47^2+45^2+...+3^2+1^2\right)\)
\(B=50^2+48^2+46^2+...+4^2+2^2-49^2-47^2-...-3^2-1^2\)
\(B=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(4^2-3^2\right)+\left(2^2-1^2\right)\)
\(B=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(4-3\right)\left(4+3\right)+\left(2-1\right)\left(2+1\right)\)
\(B=50+49+48+47+...+4+3+2+1\)
\(B=1+2+3+...+48+49+50\)
\(B=\dfrac{50-1+1}{2}.\left(1+50\right)\)
\(B=25.51\)
\(B=1275\)
\(50^2-49^2+48^2-47^2+.........+2^2-1^2\)
= \(\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+........+\left(2-1\right)\left(2+1\right)\)=\(50+49+48+47+......+2+1\)
Quy luật dãy khoảng cách là 1 đơn vị
=> Số Số Hạng là : (50 - 1) : 1 + 1 = 50 (số)
=> tổng dãy là :
\(50\cdot\dfrac{\left(50+1\right)}{2}=1275\)
bạn ơi cho mk hỏi 50+49+48+47+...+2+1 tính ở đâu ak bạn
THANKS
Bài 1:
a) \(100^2-99^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+...+2+1\)
=> tự làm tiếp :))
b) tương tự
Bài 2 :
a) \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(\left(2-1\right)A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^8-1\right)\left(2^8+1\right)\)
\(A=2^{16}-1< 2^6=B\)
b) Phân tích \(2004\cdot2006=\left(2005-1\right)\left(2005+1\right)=\left(2005^2-1\right)\)rồi áp dụng hđt thứ 3 tự làm tiếp như câu a)
Bài 3:
a) Cứ khai triển hết ra
b) \(a^2+b^2+c^2=ab+bc+ac\)
\(a^2+b^2+c^2-ab-bc-ac=0\)
Nhân 2 vào cả 2 vế được :
\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+c^2\right)=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
mà mũ 2 luôn lớn hơn hoặc bằng 0
\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow}a=b=c\left(đpcm\right)}\)
P.s: toàn bài nâng cao làm hơi ẩu tí ^^
Câu b :
\(A=\left(50^2+48^2+46^2+.........+4^2+2^2\right)-\left(49^2+47^2+45^2+.........+5^2+3^2+1^2\right)\)
\(A=\left(50^2-49^2\right)+\left(48^2-47^2\right)+.........\left(4^2-3^2\right)+\left(2^2-1^2\right)\)
\(A=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+..........+\left(4+3\right)\left(4-3\right)+\left(2+1\right)\left(2-1\right)\)
\(A=50+49+48+..........+3+2+1\)
\(A=\dfrac{50.51}{2}\)
\(\Rightarrow A=1275\)
a, \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)