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\(\text{a=1.2+2.3+3.4+......+98.99}\)
\(=1\left(1\text{+}1\right)\text{+}2\left(2\text{+}1\right)\text{+}3\left(3\text{+}1\right)\text{+}.........\text{+}98\left(98\text{+}1\right)\)
\(=1^2\text{+}1\text{+}1^2\text{+}2\text{+}3^2\text{+}3\text{+}...\text{+}98^2\text{+}98\)
\(=b\text{+}\left(1\text{+}2\text{+}3\text{+}...\text{+}98\right)\)
\(=b\text{+}\left(98\text{+}1\right).98:2\)
\(=b\text{+}4851\)
\(\Rightarrow a-b=4851\)
a) Ta có: A = 1.2 + 2.3 + 3.4 + 4.5 +....+ 98.99
\(\Rightarrow\) 3A = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 +....+ 98.99.3
\(\Rightarrow\) 3A = 1.2.3 + 2.3(4-1) + 3.4(5-2) + 4.5(6-3) +.....+ 98.99(100-97)
\(\Rightarrow\) 3A = 1.2.3 + 2.3.4 - 2.3.1 + 3.4.5 - 3.4.2 + 4.5.6 - 4.5.3 + ....+ 98.99.100 - 98.99.97
\(\Rightarrow\) 3A = 98.99.100
\(\Rightarrow\) A = \(\frac{98.99.100}{3}\) = 323400
Vậy A = 323400
b) Ta có: B = 12 + 22 + 32 +......+ 982
\(\Rightarrow\) B = 1.1 + 2.2 + 3.3 + .....+ 98.98
\(\Rightarrow\) B = 1(2-1) + 2(3-1) + 3(4-1) + .......+ 98(99-1)
\(\Rightarrow\) B = 1.2 - 1 + 2.3 - 2 + 3.4 - 3 +......+ 98.99 -98
\(\Rightarrow\) B = (1.2 + 2.3 + 3.4 +........+ 98.99) - (1 + 2 + 3 +.....+ 98)
\(\Rightarrow\) B = (1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 +....+ 98.99.3) : 3 - 4851
\(\Rightarrow\) B = [1.2.3 + 2.3(4-1) + 3.4(5-2) + 4.5(6-3) +.....+ 98.99(100-97)] : 3 - 4851
\(\Rightarrow\) B = (1.2.3 + 2.3.4 - 2.3.1 + 3.4.5 - 3.4.2 + 4.5.6 - 4.5.3 + ....+ 98.99.100 - 98.99.97) : 3 + 4851
\(\Rightarrow\) B = \(\frac{\text{ 98.99.100 }}{3}\) + 4851
\(\Rightarrow\) B = 323400 + 4851
\(\Rightarrow\) B = 328251
Vậy B = 328251
A = 1.2 + 2.3 + 3.4 + ... + 98.99
A = 1.(1 + 1) + 2.(2 + 1) + 3.(3 + 1) + ... + 98.(98 + 1)
A = 12 + 1 + 22 + 2 + 32 + 3 + ... + 982.98
A = (12 + 22 + 32 + ... + 982) + (1 + 2 + 3 + ... + 98)
A = (12 + 22 + 32 + ... + 982) + 4851 (1)
B = 12 + 22 + 32 + ... + 982 (2)
(1)(2) => A - B = 4851 ⋮ 4851
ta có: B = 12 + 22 + 32 +...+982 = 1.1 +2.2+3.3+...+98.98
=> A-B = (1.2+2.3+3.4+4.5+...+98.99) - (1.1+2.2+3.3+...+98.98)
A-B = (1.2-1.1) + (2.3-2.2) + (3.4-3.3) + (4.5-4.4) + ...+ (98.99-98.98)
A-B = 1.(2-1) + 2.(3-2) +3.(4-3) + 4.(5-4) + ...+ 98.(99-98)
A-B = 1 +2+3+4+...+98
A-B = (1+98).98:2
A -B = 4851 chia hết cho 4851
hahahahaha
xet A va B thi ta thay B la cac so binh phuong cong vao con A thi lai la cac so nhan nhau lien tiep neu de y ta thay neu lay 1*2-1=1....va tru thanh day so lien den 98
A-B=1+2+3+........................+98 co 98 so hang
A-B=(1+98)*98:2=4851
Ta có A =1.2 + 2.3 + 3.4 + ...+ 98.99
B = 1^2 + 2^2 + 3^2 +...+98^2 = 1.1+2.2+3.3+...+98.98
Suy ra: A-B= (1.2 + 2.3 + 3.4 + ...+ 98.99) - (1.1+2.2+3.3+...+98.98)
= (1.2-1.1) + (2.3-2.2) + (3.4-3.3) +...+ (98.99-98.98)
= 1(2-1) + 2(3-2) + 3(4-3) +...+ 98(99-98)
= 1.1 + 2.1 + 3.1 +...+ 98.1
= 1+ 2+ 3+...+ 98 = [98.(98+1)]/2= 98.99/2 = 4851
đúng nha
A = 1.2 + 2.3 + 3.4 + ... + 98.99
= 1.(1 + 1) + 2.(2 + 1) + 3.(3 + 1) + ... + 98.(98 + 1)
= 12 + 1 + 22 + 2 + 32 + 3 + ... + 982 +98
= (12 + 22 + 32 + ... + 982) + (1 + 2 + 3 + ... + 98)
= B + (1 + 98).98 : 2
= B + 4851
Do đó A = B + 4851 suy ra A - B = 4851
Kết luận : A - B = 4851
b)Ta chứng minh công thức \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\) (*)
Với n=1 (*) đúng
Giả sử (*) đúng với n=k, khi đó ta có
\(1^2+2^2+...+k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\) (1)
Ta chứng minh (1) đúng với n=k+1, từ (1) suy ra:
\(1^2+2^2+...+k^2+\left(k+1\right)^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)
\(=\left(k+1\right)\left(\frac{k\left(2k+1\right)}{6}+k+1\right)=\left(k+1\right)\frac{2k^2+7k+6}{6}\)
\(=\frac{\left(k+1\right)\left(2k^2+4k+3k+6\right)}{6}=\frac{\left(k+1\right)\left[2k\left(k+2\right)+3\left(k+2\right)\right]}{6}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)
Theo nguyên lí quy nạp ta có ĐPCM
Áp dụng vào bài toán ta có:
\(B=\frac{98\left(98+1\right)\left(2\cdot98+1\right)}{6}=318549\)
a)\(A=1\cdot2+2\cdot3+...+98\cdot99\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+98\cdot99\left(100-97\right)\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+98\cdot99\cdot100-97\cdot98\cdot99\)
\(3A=98\cdot99\cdot100=\frac{98\cdot99\cdot100}{3}=323400\)
C1 : B=\(\frac{1^2}{1.2}.\frac{2^2}{2.3}......\frac{98^2}{98.99}\)=\(\frac{1.1}{1.2}.\frac{2.2}{2.3}......\frac{98.98}{98.99}\)=\(\left(\frac{1.2......98}{1.2.....98}\right).\left(\frac{1.2......98}{2.3......99}\right)\)
\(1.\frac{1}{99}=\frac{1}{99}\)
C2:Đầu tiên cũng tách ra:\(1^2\)=1.1;\(2^2\)=2.2;...;\(98^2\)=98.98
Xong rút gọn ở tử và mẫu được:\(\frac{1}{2}.\frac{2}{3}.......\frac{98}{99}=\frac{1.2.....98}{2.3.....99}=\frac{1}{99}\)
Bạn thấy cách nào rễ hiểu hơn thì ghi nhé