Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: S=1(1+1)+2(1+2)+...+100(1+100)
=1+2+...+100+1^2+2^2+...+100^2
\(=\dfrac{100\cdot102}{2}+\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\)
\(=100\cdot51+\dfrac{100\cdot101\cdot201}{6}\)
=343450
b: \(A=1\cdot2\cdot3+2\cdot3\cdot4+...+100\cdot101\cdot102\)
=>\(4\cdot A=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+...+100\cdot101\cdot102\left(103-99\right)\)
=>4*A=100*101*102*103
=>A=25*101*102*103
A = \(\dfrac{3^{100}.\left(-2\right)+3^{101}}{\left(-3\right)^{101}-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2\right)+3^{100}.3}{\left(-3\right)^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2+3\right)}{3^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.1}{3^{100}.\left(-3-1\right)}\)
A = \(\dfrac{3^{100}}{3^{100}}\) . \(\dfrac{1}{-4}\)
A = - \(\dfrac{1}{4}\)
C=(1x3+3x5+...+99x101)+(2x4+4x6+...+98x100)
đặt S=1x3+3x5+...+99x101
=>6S=6x(1x3+3x5+...+99x101)
=1x3x(5+1)+3x5x(7-1)+...+97x99x(101-95)+99x101x(103-97)
=1x3x5+1x3x1+3x5x7-1x3x5+....+97x99x101-95x97x99+99x101x103-97x99x101
=1x3x1+99x101x103
=>S=(3+99x101x103):6=171650
=>C=171650+(2x4+4x6+...+98x100)
đặt A=2x4+4x6+...+98x100
=>6A=6x(2x4+4x6+...+98x100)
=>6A=2x4x6+4x6x(8-2)+...+96x98x(100-94)+98x100x(102-96)
=2x4x6+4x6x8-2x4x6+...+96x98x100-94x96x98+98x100x102-96x98x100
=98x100x102
=>A=98x100x102:6=166600
=>C=166600+171650
=>C=338250
B=2x2+4x4+6x6+...+100x100
=2x(4-2)+4x(6-2)+6x(8-2)+...+100x(102-2)
=2x4-4+4x6-8+6x8-12+...+100x102-200
=(2x4+4x6+6x8+...+100x102)-(4+8+12+...+200)
đặt A=2x4+4x6+...+98x100+100x102
=>6A=6x(2x4+4x6+...+98x100+100x102)
=>6A=2x4x6+4x6x(8-2)+...+96x98x(100-94)+98x100x(102-96)+100x102x(104-98)
=2x4x6+4x6x8-2x4x6+...+96x98x100-94x96x98+98x100x102-96x98x100+100x102x104-98x100x102
=100x102x104
=>A=100x102x104:6=176800
=>B=176800-(4+8+12+...+200)
đặt S=4+8+12+..+200
Số số hạng của S là:
(200-4):4+1=50 số
S=(200+4)x50:2=5100
=>B=176800-5100
=>B=171700
bạn phá ngoặc đi, rồi dùng quy tắc chuyển vế chuyển x ở vế bên phải sang vế bên trái rồi tính như bình thường.
vi |x+1|>0
|x+ 2|>0
.....làm như vậy đến |x+100|
suy ra 100x>0
suy ra x+1+x+2+x+3+....+x+100=101x
100x+ (100+1).100:2=101x
100x+5050=101x
5050=101x-100x
x=5050
Đặt A=\(2^{x+1}+2^{x+2}+...+2^{x+100}\)
=>\(2\cdot A=2^{x+2}+2^{x+3}+...+2^{x+101}\)
=>\(A=2^{x+101}-2^{x+1}\)
\(A=32\left(2^{101}-2\right)=2^{106}-2^6\)
=>\(2^{x+1}\left(2^{100}-1\right)=2^6\left(2^{100}-1\right)\)
=>x+1=6
=>x=5