Question

Find the n-th term of the sequence, \[10,23,60,169,494 \ldots \]

Answer 1

Hint: This problem can be solved by checking by taking the difference of the successive two terms. The process is to be repeated twice. Doing this series will be in geometric progression or arithmetic progression.

Complete step-by-step solution:
The given series is
\[10,23,60,169,494 \ldots \]
Subtracting first term from the second, seconds from third, third from the fourth, and so on.
$\left( {23 - 10} \right),\left( {60 - 23} \right),\left( {169 - 60} \right),\left( {494 - 169} \right), \ldots $
The series now becomes after subtraction as,
$13,37,109,325, \ldots $
Now again repeating the same step again, subtracting first term from the second, seconds from third, third from the fourth, and so on.
$\left( {37 - 13} \right),\left( {109 - 37} \right),\left( {325 - 109} \right), \ldots $
The terms after subtraction becomes as,
$24,72,216, \ldots \cdots \left( 1 \right)$
If we analyze the terms of equation (1), it is clear that the terms are following geometric progression in which the ratio of second term to first term and third term to second term is 3.
The ratio of second and first term
$
  r = \dfrac{{{a_2}}}{{{a_1}}} \\
   \Rightarrow r = \dfrac{{72}}{{24}} \\
   \Rightarrow r = 3 \\
 $
The ratio of second and first term
$
  r = \dfrac{{{a_3}}}{{{a_2}}} \\
   \Rightarrow r = \dfrac{{216}}{{72}} \\
   \Rightarrow r = 3 \\
 $
The n-th term of this GP is given by after 2 successive subtractions,
$
  {u_n} = a{r^{n - 1}} + bn + c \\
   \Rightarrow {u_n} = a{.3^{n - 1}} + bn + c \cdots \left( 2 \right) \\
 $
Put $n = 1$ , in equation (2),
$
  {u_1} = a{\left( 3 \right)^{1 - 1}} + b\left( 1 \right) + c \\
   \Rightarrow 10 = a + b + c \\
   \Rightarrow a + b + c = 10 \cdots \left( 3 \right) \\
 $
Put $n = 2$ , in equation (2),
$
  {u_1} = a{\left( 3 \right)^{2 - 1}} + b\left( 2 \right) + c \\
   \Rightarrow 10 = 3a + 2b + c \\
   \Rightarrow 3a + 2b + c = 23 \cdots \left( 4 \right) \\
 $
Put $n = 3$ , in equation (2),
$
  {u_1} = a{\left( 3 \right)^{3 - 1}} + b\left( 3 \right) + c \\
   \Rightarrow 60 = 9a + 3b + c \\
   \Rightarrow 9a + 3b + c = 60 \cdots \left( 5 \right) \\
 $
There are three equations and three unknowns.
Put the value of equation (1) as $a = 10 - b - c$ in equation (4) and (5) to obtain the two equations in terms of $b$ and$c$.
Equation (4) becomes as ,
$
  3\left( {10 - b - c} \right) + 2b + c = 23 \\
   \Rightarrow 30 - 3b - 3c + 2b + c = 23 \\
   \Rightarrow b + 2c = 7 \cdots \left( 6 \right) \\
 $
Equation (5) becomes as ,
$
  9\left( {10 - b - c} \right) + 3b + c = 60 \\
   \Rightarrow 90 - 9b - 9c + 3b + c = 60 \\
   \Rightarrow 6b + 8c = 30 \\
   \Rightarrow 3b + 4c = 15 \cdots \left( 7 \right) \\
 $
Multiply equation (6) by 3,
$
  3\left( {b + 2c} \right) = 3 \times 7 \\
   \Rightarrow 3b + 6c = 21 \cdots \left( 8 \right) \\
 $
Subtracting equation (7) from equation (8),
$
  \left( {3b + 6c} \right) - \left( {3b + 4c} \right) = 21 - 15 \\
   \Rightarrow 2c = 6 \\
   \Rightarrow c = 3 \\
 $
Substituting the value of c in equation (7), to obtain value of $b$
$
  3b + 4\left( 3 \right) = 15 \\
   \Rightarrow 3b + 12 = 15 \\
   \Rightarrow 3b = 3 \\
   \Rightarrow b = 1 \\
 $
Substitute the value of b and c in equation (3), to obtain value of $b$
$
  a + 1 + 3 = 10 \\
   \Rightarrow a = 6 \\
 $
Now, the n-th term is given by substituting the value of $a,b$ and $c$ in equation (2),
${u_n} = {6.3^{n - 1}} + n + 3$ where, n=1,2,3,4………

Note: The important thing is to analyze that the series becomes geometric progression after two subtraction of the successive terms. Since the 2 times subtraction is performed, two more terms $b$ and $c$ are added as ${u_n} = a{r^{n - 1}} + bn + c$.