Question

Find the nth term and the sum of n terms of the series\[8,16,0, - 64, - 200, - 432, \ldots \]

Answer 1

Hint: Compute the successive differences between each pair and repeat the step until you get the differences as constant ignoring the signs in the last step. Define the nth term as \[{t_n} = a + bn + c{n^2} + d{n^3}\]. Solve the system of equations obtained using the first four terms of the series and find the values of a, b, c, and d. This gives the nth term. Use the formula for sum of n terms ${S_n} = \sum {{t_n}} $and find the answer using the following identities:
1) the sum of squares of first n natural numbers is given by\[\dfrac{{n(n + 1)(2n + 1)}}{6}\]
2) the sum of cubes of first n natural numbers is given by \[{[\dfrac{{n(n + 1)}}{2}]^2}\]

Complete step by step solution:
There are two parts to the given question.
First part: Find the nth term of the series\[8,16,0, - 64, - 200, - 432, \ldots \]
Second part: Find the sum of n terms of the series\[8,16,0, - 64, - 200, - 432, \ldots \]
Let us look at the first part.
Finding the nth term is equivalent to finding a general formula to calculate any term of the series.
Once the nth term is known, finding the sum requires just an application of a formula.
Consider the difference between the consecutive terms of the given series:
\[1)8 - 16,16 - 0,0 - \left( { - 64} \right), - 64 - \left( { - 200} \right), - 200 - \left( { - 432} \right), \ldots \]
\[ \Rightarrow - 8,16,64,136,232, \ldots \]
Repeat this step until we get equal differences.
\[\begin{array}{*{20}{l}}
  {2) - 8 - 16,16 - 64,136 - 64,232 - 136, \ldots } \\
  { \Rightarrow - 24, - 48,72,96, \ldots }
\end{array}\]
\[\begin{array}{*{20}{l}}
  {3) - 24 - \left( { - 48} \right),\left( { - 48} \right) - 72,72 - 96, \ldots } \\
  { \Rightarrow 24, - 24, - 24, \ldots }
\end{array}\]
If we ignore the signs, we can see that the differences are constant at the third step.
Let the nth term be denoted by\[{t_n}\]and is defined as\[{t_n} = a + bn + c{n^2} + d{n^3}\]
As the third differences were constant, we define the nth term only till the third power.
Thus, by substituting the value of n = 1, 2, 3, and so on, we have the first term \[{t_1} = a + b + c + d = 8\]
The second term\[{t_2} = a + 2b + 4c + 8d = 16\]
The third term\[{t_3} = a + 3b + 9c + 27d = 0\]
The fourth term\[{t_4} = a + 4b + 16c + 64d = - 64\]
Thus, we have a system of 4 linear equations in 4 unknowns.
We can solve them simultaneously.
\[{t_2} - {\text{ }}{t_1}\]gives us$b + 3c + 7d = 8........(1)$
\[{t_3} - {\text{ }}{t_2}\]gives us$b + 5c + 19d = - 16..........(2)$
Further, we subtract (1) from (2) to obtain$2c + 12d = - 24..........(3)$
Now, \[{t_4} - {\text{ }}{t_3}\]gives us$b + 7c + 37d = - 64..........(4)$
Subtracting (2) from (4), we get$2c + 18d = - 48..........(5)$
(5) - (3) gives us$d = - 4$
Substituting$d = - 4$in (3) gives$c = 12$
Using the values $c = 12$and$d = - 4$in\[{t_1} = a + b + c + d = 8\], we get$a = b = 0$
Thus, the nth term\[{t_n} = a + bn + c{n^2} + d{n^3}\]becomes\[{t_n} = 12{n^2} - 4{n^3}\]
Hence the answer to the first part is \[{t_n} = 12{n^2} - 4{n^3}\]
Now, to calculate the sum of the first n terms of the series, consider the following formula:
${S_n} = \sum {{t_n}} $where Sn is the sum of the first n terms of the series.
\[
   \Rightarrow {S_n} = \sum {{t_n}} = \sum {(12{n^2} - 4{n^3})} \\
   \Rightarrow {S_n} = \sum {12{n^2} - \sum {4{n^3}} } \\
   \Rightarrow {S_n} = 12\sum {{n^2}} - 4\sum {{n^3}} \\
 \]
We know that the sum of squares of first n natural numbers is given by\[\dfrac{{n(n + 1)(2n + 1)}}{6}\]
Also, the sum of cubes of first n natural numbers is given by \[{[\dfrac{{n(n + 1)}}{2}]^2}\]
\[
   \Rightarrow {S_n} = 12[\dfrac{{n(n + 1)(2n + 1)}}{6}] - 4{[\dfrac{{n(n + 1)}}{2}]^2} \\
   \Rightarrow {S_n} = 2n(n + 1)(2n + 1) - {n^2}{(n + 1)^2} \\
   \Rightarrow {S_n} = n(n + 1)[2(2n + 1) - n(n + 1)] \\
   \Rightarrow {S_n} = n(n + 1)(4n + 2 - {n^2} - n) \\
   \Rightarrow {S_n} = n(n + 1)( - {n^2} + 3n + 2) \\
   \Rightarrow {S_n} = - n(n + 1)({n^2} - 3n - 2) \\
 \]
Hence the required sum is\[{S_n} = - n(n + 1)({n^2} - 3n - 2)\].
Note: This method of finding the nth term is applicable only if the differences between the successive terms become constant after repeated steps.