Question

Given the sequence: 1;x; 6;y:15 ... The sum of the first four terms is 20 and the second differences are equal 1.1 Find a formula for the nth term Tn​

Answer 1

Explanation:

Here is a methode more general:

`\Delta_(1,n)=u_(n+1)-u_(n)\\\\\Delta_(1,n+1)=u_(n+2)-u_(n+1)\\\\\\\Delta_(2,n)=\Delta_(1,n+1)-\Delta_(1,n)=u_(n+2)-u_(n+1)-(u_(n+1)-u_(n))\\=u_(n+2)-2*u_(n+1)+u_(n)=1\\\\\Delta_(2,n+1)=u_(n+3)-2*u_(n+2)+u_(n+1)=1\\\\\Delta_(2,n+1)-\Delta_(2,n)=u_(n+3)-3*u_(n+2)+3u_(n+1)-u_n=0\\\\Caracteristic\ equation: \ r^3-3r^2+3r-1=0=(r-1)^3\\\\u_n=k_1*1^n+k_2*n*1^n+k_3*n^2*1^n\\\\`

`\left\{\begin{array}{ccc}u_1&=&1\\u_2&=&x\\u_3&=&6\\\end {array}\right.\\\\\\\left\{\begin{array}{ccc}k_1+k_2+k_3&=&1\\k_1+2*k_2+4*k_3&=&x\\k_1+3*k_2+9*k_3&=&6\\\end {array}\right.\\\\\\\left\{\begin{array}{ccc}k_1&=&-3x+9\\\\k_2&=&4x-(23)/(2)\\\\k_3&=&-x+(7)/(2)\\\end {array}\right.\\u_n=-3x+9+(4x-(23)/(2))*n+(-x+(7)/(2))*n^2\\\\u_1=-3x+4x-x+9-(23)/(2) +(7)/(2) =1\\u_2=-3x+8x-4x+9-23 +14=x\\u_3=-3x+12x-9x+9-(69)/(2) +(63)/(2) =6\\`

`u_4=-3x+16x-16x+9-46 +56 =-3x+19=y\\u_5=-3x+20x-25x+9-(115)/(2) +(175)/(2) =-8x+39=15\\\\\Longrightarrow\ x=3\\k_1=-9+9=0\\k_2=4*3-(23)/(2) =(1)/(2)\\k_3=-3+(7)/(2) =(1)/(2)\\\\\\u_n=0+(n)/(2)+(n^2)/(2)=(n*(n+1))/(2)\\\\Sequence\ is\ 1;3;6;10;15;...\\`

Answer 2

9514 1404 393

Answer:

Tn = (n/2)(n +1)

Explanation:

The first differences are ...

x -1, 6 -x, y -6, 15 -y

Then the second differences are ...

(6 -x) -(x -1) = 1 ⇒ 7 -2x = 1 ⇒ x = 3

(y -6) -(6 -x) = 1 ⇒ x + y = 13 ⇒ y = 10

(15 -y) -(y -6) = 1 ⇒ 21 -2y = 1 ⇒ y = 10

Then the sequence is ...

1, 3, 6, 10, 15, ...

which is the sequence of triangle numbers. The formula for the general term is ...

Tn = (n/2)(n +1)