D = 1.2 + 2.3 + 3.4 + ... + 99.100

3D = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3

3D = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)

3D = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

3D = 99.100.101

D = 99.100.101 : 3 = ...

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

a)N=1.2+2.3+3.4+4.5+......+99.100

3N=1.2.3+2.3.3+3.3.4+3.4.5+......+3.99.100

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=999900

N=333300

b)P=10.11+11.12+12.13+....+29.30

3P=10.11.3+11.12.3+....+29.30.3

3P=10.11.12-9.10.11+11.12.13-10.11.12+.........+29.30.31-28.29.30

3P=29.30.31-9.10.11=25980

P=8660

R=1²+2²+3²+....+100²

R=1.(2-1)+2.(3-1)+.........+100.(101-1)

R=1.2-1+2.3-2+3.4-3+.....+100.101-100

R=(1.2+2.3+3.4+..........+100.101)-(1+2+3+.........+99+100)

\(R=\left(N+100.101\right)-\frac{100.101}{2}\)

R=343400-5050=338350

S=100.1+2.(100-1)+3.(100-2)+......+99.(100-98)+100(100-99)

S=100.1+100.2-1.2+100.3-2.3+.......+99.100-98.99+100.100-99.100

S=100(1+2+...+100)-(1.2+2.3+....+99.100)

S=100.(100.101:2)-N

S=505000-333300=171700

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!!! ~

Tổng của B ;

\(SUM\left(B\right)=\sum\limits^{100}_{x=1}\left(x^2\right)=338350\)

Tổng của C :

\(SUM\left(C\right)=\sum\limits^{200}_{x=101}\left(x^2\right)=2348350\)

1

no

CÂU A NHÂN A VS 3 RỒI TRỪ ĐI A RỒI LẤY 1 SỐ NHÂN VS 1 TỔNG

CÂU B , C DỰA VÀO CÂU A

CHÚC BN THÀNH CÔNG !!

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