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ta tách một số n như sau
\(n^2=n\cdot n=n\left(n-1+1\right)=n\left(n-1\right)+n\)
=> \(1^2=1\cdot0+1\)
\(2^2=2\cdot1+2\)
\(3^2=3\cdot2+3\)
.....
\(100^2=100\cdot99+100\)
=> \(B=\left(1\cdot0+2\cdot1+3\cdot2+\cdots+100\cdot99\right)+\left(1+2+\cdots+100\right)\)
nhân 3 vào ngoặc 1
=> \(=\frac{99.100.101}{3}=333300\) ( mik hơi lười vt bạn tự tra nha:v)
\(\Rightarrow B=333300+5050=338350\)
b) Áp dụng công thức tính tổng bình phương:
\(1^2+\cdots+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\) ( từ câu a)
tổng từ \(1^2\) đến \(200^2\) là:
\(\frac{200.201.401}{6}=2686700\)
tổng từ \(1^2\) đến \(100^2\) là: \(338350\)
=>C=\(2686700-338350=2348350\)
c) tách số như sau:
\(1\cdot3=1\cdot2+1\)
\(2\cdot4=2\cdot3+2\)
...
\(100\cdot102=100\cdot101+100\)
=> \(S=\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)+\left(1+2+3+\cdots+100\right)\)
=> \(S=\frac{100.101.102}{3}+\frac{100.101}{2}\)
=> \(S=343400+5050=348450\)
d) \(1\cdot100=1\cdot101-1^2\)
\(2\cdot99=2\cdot101-2^2\)
...
\(100\cdot1=100\cdot101-100^2\)
=>T=\(\left(1\cdot101+2\cdot101+\cdots+100\cdot101\right)-\left(1^2+2^2+\cdots+100^2\right)\)
\(T=101\left(1+2+3+\cdots+100\right)-B\) (B là tổng ở câu a) nha)
T=\(101\cdot5050-338350\)
\(T=171700\)
e) => \(4E=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+\cdots+98\cdot99\cdot100\cdot4\)
\(4E=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+\cdots+98\cdot99\cdot100\cdot\left(101-97\right)\)
\(4E=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+\cdots+98\cdot99\cdot100\cdot101-97\cdot98\cdot99\cdot100\)
=> \(4E=98\cdot99\cdot100\cdot101\)
=> \(E=24497550\)
3N = 1.2.3+2.3(4-1)+3.4.(5-2)+.+99.100.(101-98)
3N = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.+99.100.101-98.99.100
3N = 99.100.101
3N=33.100.101=333300
b)
tổng này có 99-10+1=90 (số hạng):
10,11 + 11,12 + 12,13 +............+ 98,99 + 99,100 =
10,100 + 11,11 + 12,12 + .......... + 98,98 + 99,99 =
(10,10 + 99,99) x 90 : 2 = 4954,05
c)
R=1.(2-1)+2.(3-1)+.....+100.(101-1)
=1.2-1.1+2.3-1.2+......+100.101-1.100
=(1.2+2.3+.....+99.100+100.101)-(1+2+3+...+100)
=[1.2.3+2.3.(4-1)+........100.101.(102-99)]:3+[(100+1).100:2]
(tổng trên chia cho 3 nên cuối cùng chia 3)
=(1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....100.101.102-99.100.101):3+5050
=(100.101.102) :3 +5050
=348450
d)=1.100+2.(100-1)+.....+100.(100-99)
=1.100+2.100-1.2+3.100-2.3+........+100.100-99.100
=100.(1+2+3+.......+100)-(1.2+2.3+3.4+....+99.100)
=100.\(\frac{101.100}{2}-\frac{99.100.101}{3}\) =505000-333300=171700
p/s mỏi tay, bấm mình nhé
\(\frac{1+\left[1+2\right]+\left[1+2+3\right]+...+\left[1+2+3+...+100\right]}{100.1+99.2+98.3+...+2.99+1.100}=\frac{1.2:2+2.3:2+3.4:2+...+100.101:2}{100.1+99.2+98.3+...+2.99+1.100}\)
\(=\frac{\frac{1}{2}\left[1.2+2.3+3.4+...+100.101\right]}{100.1+99.2+98.3+...+2.99+1.100}=\frac{\frac{1}{2}\cdot\frac{1}{3}\left[1.2.3-0.1.2+2.3.4-1.2.3+...+100.101.102-99.100.101\right]}{1.100+2.100-1.2+3.100-2.3+...+100.100-99.100}\)
\(=\frac{\frac{1}{6}\cdot100.101.102}{100\left[1+2+3+...+100\right]-\left[1.2+2.3+...+99.100\right]}=\frac{171700}{100\cdot\frac{100.101}{2}-\frac{99.100\cdot101}{3}}\)
\(=\frac{171700}{505000-333300}=\frac{171700}{171700}=1\)
AI THẤY ĐÚNG NHỚ ỦNG HỘ NHÉ
\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+100\right)}{100.1+99.2+...+1.100}\)
\(\frac{1+1+2+1+2+3+...+1+2+...+100}{100.1+99.2+...+1.100}\)
\(=\frac{1.100+2.99+3.98+...+100.1}{100.1+99.2+...+1.100}\)
\(=1\)
Giải:
a, \(B=1^2+2^2+3^2+...+99^2+100^2.\)
\(B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+99\left(100-1\right)+100\left(101-1\right).\)
\(B=1.2-1.1+2.3-1.2+3.4-1.3+...+99.100-1.99+100.101-1.100.\)
\(B=\left(1.2+2.3+3.4+...+99.100+100.101\right)-\left(1+2+3+...+100\right).\)
\(B=\dfrac{\left[1.2.3+2.3\left(4-1\right)+3.4\left(5-2\right)+...+100.101\left(102-99\right)\right]}{3}+\dfrac{100\left(100+1\right)}{2}.\)
\(B=\dfrac{\left(1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101\right)}{3}+5050.\)
\(B=\dfrac{100.101.102}{3}+5050.\)
\(B=343400+5050=348450.\)
Vậy \(B=348450.\)
\(C=...\) (làm tương tự con \(B\)).
\(D=...\) (hình như đề sai).
\(T=1.100+2.99+3.98+...+99.2+100.1.\)
\(T=1.100+2.\left(100-1\right)+3.\left(100-2\right)+...+99\left(100-98\right)+100\left(100-99\right).\)
\(T=1.100+100.2+1.2+100.3+2.3+...+100.99+98.99+100.100+99.100.\)
\(T=100\left(1+2+3+...+100\right)-\left(1.2+2.3+3.4+...+99.100\right).\)
\(T=100.\dfrac{100.101}{2}-\dfrac{99.100.101}{3}.\)
\(T=100.5050-333300.\)
\(T=505000-333300=171700.\)
Vậy \(T=171700.\)
\(S=1.2.3+2.3.4+3.4.5+...+98.99.100.\)
\(4S=4\left(1.2.3+2.3.4+3.4.5+...+98.99.100\right).\)
\(4S=1.2.3.4+2.3.4.4+3.4.5.4+...+98.99.100.4.\)
\(4S=1.2.3\left(5-1\right)+2.3.4\left(6-2\right)+...+98.99.100\left(101-97\right).\)
\(4S=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100.\)
\(4S=\left(1.2.3.4-1.2.3.4\right)+\left(2.3.4.5-2.3.4.5\right)+...+\left(97.98.99.100-97.98.99.100\right)+98.99.100.101.\)
\(4S=0+0+...+0+98.99.100.101.\)
\(4S=98.99.100.101.\)
\(4S=97990200.\)
\(\Rightarrow S=\dfrac{97990200}{4}=24497550.\)
Vậy \(S=24497550.\)
~ Học tốt!!! ~