\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)

\(=6n^2+30n+n+5-\left(6n^2-3n+10n-5\right)\)

\(=6n^2+31n+5-6n^2-7n+5\)

\(=24n+10\)

\(=2\left(12n+5\right)\) chia hết cho 2

=> \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)chia hết cho 2 (Đpcm)

\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)

\(=6n^2+30n+n+5-\left(6n^2-3n+10n-5\right)\)

\(=6n^2+31n+5-6n^2-7n+5\)

\(=24n+10\)

\(=2\left(12n+5\right)⋮2\)

\(\Rightarrow\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)⋮2\) ( đpcm )

S=(3^0+1/2)+(3^1/2+1/2)+(3^2/2+1/2)+....+(3^n-1/2+1/2)

=n*1/2+1/2*(3^0+3^1+3^2+...+3^n-1)

=n^2/2+(3^n-1/4)=3^n+2-1/4

~~~~~~~~~~~~~~~~~~~~~

\(S=1+2+5+14+....+\frac{3^{x-1}+1}{2}\)

\(=\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+.....+\frac{3^{x-1}+1}{2}\)

\(=\frac{\left(3^0+1\right)+\left(3^1+1\right)+\left(3^2+1\right)+.....+\left(3^{x-1}+1\right)}{2}\)

\(=\frac{\left(1+3+3^2+.....+3^{x-1}\right)+x}{2}\)

Đặt \(A=1+3+3^2+....+3^{x-1}\)

\(3A-A=\left(3+3^2+....+3^x\right)-\left(1+3+....+3^{x-1}\right)\)

\(2A=3^x-1\Rightarrow A=\frac{3^x-1}{2}\)

\(\Rightarrow S=\frac{\frac{3^x-1}{2}+x}{2}\)

áp dụng quy tắc 

số số hạng= (số cuối-số đầu) chí cho khoảng cách rồi cộng với 1

Tổng=(số đầu +số cuối ) nhân với số số số hạng rồi chia cho 2

Bài 1:

a) Ta có: \(x=7\Rightarrow8=x+1\)

Thay vào ta được:

\(A=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)

\(A=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-...-x^3-x^2+x^2+x-5\)

\(A=x-5\)

\(A=7-5=2\)

Vậy khi x = 7 thì A = 2

2)

a) 2n+5 chia het cho n-1 

=> 2(n-1) +7 chia het cho n-1 

=: n-1 thuoc uoc cua 7 den day ke bang la xong. 

may cau con lai lam tuong tu

dài quá ko mún làm

khó quá