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

9514 1404 393

Answer:

  Tn = (n/2)(n +1)

Step-by-step 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)

Answer 2

Answer:

Step-by-step explanation:

Here is a methode more general:

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

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

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