Thay a = -1 , b=1 vào biểu thức A 

=> A = 5.(-1)^3.1^8 = - 5

Thay a = -1 , b= 2 vào biểu thức B

=>B = -9.(-1)^4 . 2^2 = - 36

Ta có : 

C = ax + ay + bx + by = a(x+y) + b(x+y) = (x+y)(a+b)

Thay a+b = - 3 , x+y = 17 vào biểu thức C

C = ( -3)(17) = -51

Ta sẽ chứng minh \(1+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)(*).

Với \(n=1\)thì: \(\frac{1\left(1+1\right)\left(2.1+1\right)}{6}=1\)do đó (*) đúng với \(n=1\).

GIả sử (*) đúng với \(n=k\ge1\), tức là \(1+2^2+3^2+...+k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\).

Ta sẽ chứng minh (*) đúng với \(n=k+1\), tức là \(1+2^2+3^2+...+k^2+\left(k+1\right)^2=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\).

Thật vậy, ta có: 

\(1+2^2+3^2+...+k^2+\left(k+1\right)^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\frac{6\left(k+1\right)^2}{6}\)

\(=\frac{\left(k+1\right)\left(2k^2+k+6k+6\right)}{6}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)

Suy ra (*) đúng với \(n=k+1\).

Theo nguyên lí quy nạp toán học, (*) đúng với \(n\inℕ\).

Vậy \(1+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\).

Ta có A = 1.1 + 2.2 + 3.3 + ... + n.n 

= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + ... + n.(n + 1 - 1) 

= 1.2 + 2.3 + 3.4 + .... + n.(n + 1) - (1 + 2 + 3 + ... + n) 

= 1.2 + 2.3 + 3.4 + .... + n.(n + 1) - n(n + 1) : 2

Đặt B = 1.2 + 2.3 + 3.4 + .... + n(n + 1)

=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + .... + n.(n + 1).3

= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + .... + n.(n + 1).[(n + 2) - (n - 1)]

= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + n(n + 1)(n + 2) - (n - 1)n(n + 1)

= n(n + 1)(n + 2)

=> B = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

Khi đó \(A=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}=n\left(n+1\right)\left(\frac{n+2}{3}-\frac{1}{2}\right)\)

\(=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\)

=> 1 - \(\frac{1}{32}\)

= \(\frac{32}{32}-\frac{1}{32}\)

= \(\frac{31}{32}\)

=\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\)

=\(1-\left(\frac{1.16}{2.16}\right)-\left(\frac{1.8}{4.8}\right)-\left(\frac{1.4}{8.4}\right)\left(\frac{1.2}{16.2}\right)-\frac{1}{32}\)

=\(1-\frac{16}{32}-\frac{8}{32}-\frac{4}{32}-\frac{2}{32}-\frac{1}{32}\)

=\(1-\frac{1}{32}\)

=\(\frac{31}{32}\)

cac ban con cach giai nao khac khong

\(D=\left(2^9.3+2^9.5\right)-2^{12}\)

\(D=2^9.\left(3+2\right)-2^{12}\)

\(D=2^9.5-2^{12}\)

\(D=512.5-4096\)

\(D=2560-4096\)

\(D=-1536\)

\(\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+100^2\right).\left(65.111-13.15.37\right)\)

\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+100^2\right).\left(7215-7215\right)\)

\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+100^2\right).0\)

\(=0\)

D=(2⁹.3+2⁹.5)-2¹²

D=((2⁹.(3+5))-2¹²

D=(2⁹.8)-2¹²

D=(2⁹.2³)-2¹²

D=2⁹⁺³-2¹²

D=2¹²-2¹²

D=0

(1+2+3+...+100).(1²+2²+3²+....+100²).(65.111-13.15.37)

=(1+2+3+...+100).(1²+2²+3²+....+100²).7215-7215

=(1+2+3+...+100).(1²+2²+3²+....+100²).0

=0

C=2¹⁰-2

C=2⁹⁺¹-2

C=2⁹.2-2

C=2.(2⁹-1)

C=2.(512-1)\

C=2.511

C=1022

F=1+3¹+3²+3³+......+3¹⁰⁰

F=3+3¹+3²+3³+......+3¹⁰⁰

3F=3.(3+3¹+3²+3³+......+3¹⁰⁰)

3F=3²+3²+3³+3⁴+......+3¹⁰⁰

3F-F=3¹⁰⁰+3²-3-3

2F=3¹⁰⁰+9-3-3

F=\(\frac{3^{100}+3}{2}\)

Chúc bn học tốt

Tính G hả bạn

18/35

b.\(\dfrac{18}{35}\)