Question

Show that:

a) The 2nd differences in the following row are constant and calculate:
b)Tn
4; 7; 12; 19;....
-1; 2; 7; 14;....
-2;-8; -18; -32; ...​

Answer 1

Hello,

Explanation:

I am going to explain the method with the 1th:

`\begin{array}{ccccc}n & u_n&u_(n+1)-u_(n)&u_(n+2)-2u_(n+1)+u_(n)&v_(n)\\1&4&&6\\2&7&3&9\\3&12&5&2&14\\4&19&7&2&21\\...&&&\\\end{array}\\\boxed{u_(n+2)=2u_(n+1)-u_(n)+2}\\`

`Let\ say\\v_(n)=u_(n)+2\\v_(n+2)=u_(n+2)+2=(2u_(n+1)-u_(n)+2)+2=2(v_(n+1)-2)-(v{n}-2)+4\\v_(n+2)=2v_(n+1)-v{n}+2\ (1)\\v_(n+3)=2v_(n+2)-v{n+1}+2\ (2)\\(2)-(1)==> \boxed{v_(n+3)=3v_(n+2)-3v_(n+1)+v_n}\\`

`Caracteristic\ equation:\\P(r)=r^3-3r^2+3r-1=0\\P(r)=(r-1)^3\\v_n=\alpha+\beta*n+\gamma*n^2\\v_1=6 ==> \alpha+\beta*1+\gamma*1=6\\u_2=9 ==> \alpha+\beta*2+\gamma*4=9\\u_3=14 ==> \alpha+\beta*3+\gamma*9=14\\`

`\begin{bmatrix}1&1&1\\1&2&4\\1&3&9\end{bmatrix}*\begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}=\begin{bmatrix}6\\9\\14 \end{bmatrix}\\\\\\\begin{bmatrix}1&1&1&6\\1&2&4&9\\1&3&9&14\end{bmatrix}\\\\\\\begin{bmatrix}1&1&1&6\\0&2&3&3\\0&2&8&8\end{bmatrix}\\\\\\\begin{bmatrix}1&1&1&6\\0&1&3&3\\0&1&4&4\end{bmatrix}\\\\`

`\begin{bmatrix}1&1&1&6\\0&1&3&3\\0&0&1&1\end{bmatrix}\\\\\\\begin{bmatrix}1&1&1&6\\0&1&0&0\\0&0&1&1\end{bmatrix}\\\\\\\begin{bmatrix}1&0&0&5\\0&1&0&0\\0&0&1&1\end{bmatrix}\\`

`\alpha=5\\\beta=0\\\gamma=1\\\boxed{v_n=5+0*n+1*n^2}\\\boxed{u_n=5+0*n+1*n^2-2}\\\begin{array}{ccccc}n & u_n\\1&5+1-2=4\\2&5+4-2=7\\3&5+9-2=12\\4&5+16-2=19\\...&...\\\end{array}\\`

Answer 2

• {4, 7, 12, 19, … }

has 1st differences

7 - 4 = 3

12 - 7 = 5

19 - 12 = 7

and 2nd differences

5 - 3 = 2

7 - 5 = 2

• {-1, 2, 7, 14, …}

1st differences:

2 - (-1) = 3

7 - 2 = 5

14 - 7 = 7

2nd differences:

5 - 3 = 2

7 - 5 = 2

• {-2, -8, -18, -32, …}

1st differences:

-8 - (-2) = -6

-18 - (-8) = -10

-32 - (-18) = -14

2nd differences:

-10 - (-6) = -4

-14 - (-10) = -4