Question

If \[\sum\limits_{r = 1}^n {r\left( {r + 1} \right)\left( {2r + 3} \right)} = a{n^4} + b{n^3} + c{n^2} + dn + e\] then,
A. $a + c = b + d$
B. \[e = 0\]
C. $a,b - \dfrac{2}{3},c - 1$ are in A.P.
D. $\dfrac{c}{a}$ is an integer

Answer 1

Hint: We’ll approach the solution by finding the values of a, b, c, d, and e by evaluating the left-hand side of the equation using some the well-known formulas i.e.
\[
  \sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2} \\
  \sum\limits_{r = 1}^n {{r^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \\
  \sum\limits_{r = 1}^n {{r^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} \\
  \]
Then we’ll compare our result with the right-hand side to get the required answer.

Complete step by step answer:

Given data: \[\sum\limits_{r = 1}^n {r\left( {r + 1} \right)\left( {2r + 3} \right)} = a{n^4} + b{n^3} + c{n^2} + dn + e\]
On solving the left-hand side of the equation \[\sum\limits_{r = 1}^n {r\left( {r + 1} \right)\left( {2r + 3} \right)} = a{n^4} + b{n^3} + c{n^2} + dn + e\]
\[ = \sum\limits_{r = 1}^n {r\left( {r + 1} \right)\left( {2r + 3} \right)} \]
\[ = \sum\limits_{r = 1}^n {\left( {{r^2} + r} \right)\left( {2r + 3} \right)} \]
On further expanding we get,
\[ = \sum\limits_{r = 1}^n {\left( {2{r^3} + 3{r^2} + 2{r^2} + 3r} \right)} \]
\[ = \sum\limits_{r = 1}^n {\left( {2{r^3} + 5{r^2} + 3r} \right)} \]
It is well known that,
\[\sum\limits_{r = 1}^n {(A + B + C} ) = \sum\limits_{r = 1}^n A + \sum\limits_{r = 1}^n B + \sum\limits_{r = 1}^n C \]
\[ \Rightarrow \sum\limits_{r = 1}^n {\left( {2{r^3} + 5{r^2} + 3r} \right)} = \sum\limits_{r = 1}^n {2{r^3}} + \sum\limits_{r = 1}^n {5{r^2}} + \sum\limits_{r = 1}^n {3r} \]
\[ \Rightarrow 2\sum\limits_{r = 1}^n {{r^3}} + 5\sum\limits_{r = 1}^n {{r^2}} + 3\sum\limits_{r = 1}^n r \]
Now, as we all know
\[
  \sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2} \\
  \sum\limits_{r = 1}^n {{r^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \\
  \sum\limits_{r = 1}^n {{r^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} \\
  \]
So, We’ll have
\[2\sum\limits_{r = 1}^n {{r^3}} + 5\sum\limits_{r = 1}^n {{r^2}} + 3\sum\limits_{r = 1}^n r = 2{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} + 5\left( {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) + 3\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)\]
\[ = 2\left( {\dfrac{{{{\left( {{n^2} + n} \right)}^2}}}{4}} \right) + 5\left( {\dfrac{{\left( {{n^2} + n} \right)\left( {2n + 1} \right)}}{6}} \right) + 3\left( {\dfrac{{{n^2} + n}}{2}} \right)\]
After expanding the second term by opening the brackets
\[ = \dfrac{1}{2}{\left( {{n^2} + n} \right)^2} + \dfrac{5}{6}\left( {2{n^3} + {n^2} + 2{n^2} + n} \right) + \dfrac{3}{2}\left( {{n^2} + n} \right)\]
Using \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]in the first term,
\[ = \dfrac{1}{2}\left( {{n^4} + {n^2} + 2{n^3}} \right) + \dfrac{5}{6}\left( {2{n^3} + 3{n^2} + n} \right) + \dfrac{3}{2}\left( {{n^2} + n} \right)\]
Now, separating the terms with respect to exponents we get,
\[ = \dfrac{1}{2}{n^4} + 2{n^3}\left( {\dfrac{1}{2} + \dfrac{5}{6}} \right) + {n^2}\left( {\dfrac{1}{2} + 3\left( {\dfrac{5}{6}} \right) + \dfrac{3}{2}} \right) + n\left( {\dfrac{5}{6} + \dfrac{3}{2}} \right)\]
On further simplification we get,
\[ = \dfrac{1}{2}{n^4} + 2{n^3}\left( {\dfrac{{3 + 5}}{6}} \right) + {n^2}\left( {\dfrac{1}{2} + \dfrac{5}{2} + \dfrac{3}{2}} \right) + n\left( {\dfrac{{5 + 9}}{6}} \right)\]
\[ = \dfrac{1}{2}{n^4} + \dfrac{8}{3}{n^3} + \dfrac{9}{2}{n^2} + \dfrac{{14}}{6}n\]
\[ = \dfrac{1}{2}{n^4} + \dfrac{8}{3}{n^3} + \dfrac{9}{2}{n^2} + \dfrac{7}{3}n\]
Therefore we now have
\[\sum\limits_{r = 1}^n {r\left( {r + 1} \right)\left( {2r + 3} \right)} = \dfrac{1}{2}{n^4} + \dfrac{8}{3}{n^3} + \dfrac{9}{2}{n^2} + \dfrac{7}{3}n\]
On computing this with the given equation, \[\sum\limits_{r = 1}^n {r\left( {r + 1} \right)\left( {2r + 3} \right)} = a{n^4} + b{n^3} + c{n^2} + dn + e\] , we get,
$
  a = \dfrac{1}{2} \\
  b = \dfrac{8}{3} \\
  c = \dfrac{9}{2} \\
  d = \dfrac{7}{3} \\
  e = 0 \\
  $
Option(B) is correct
Now,
\[
  a + c = \dfrac{1}{2} + \dfrac{9}{2} \\
   = \dfrac{{10}}{2} \\
   = 5 \\
  b + d = \dfrac{8}{3} + \dfrac{7}{3} \\
   = \dfrac{{15}}{3} \\
   = 5 \\
  \]
Since, ${\text{a + c = b + d = 5}}$
Option(A) is correct
\[
  a,b - \dfrac{2}{3},c - 1 = \dfrac{1}{2},\dfrac{8}{3} - \dfrac{2}{3},\dfrac{9}{2} - 1 \\
   = \dfrac{1}{2},2,\dfrac{7}{2} \\
  \]
Now checking for common difference,
\[
  b - \dfrac{2}{3} - a = 2 - \dfrac{1}{2} \\
   = \dfrac{{4 - 1}}{2} \\
   = \dfrac{3}{2}{\text{ }}and \\
  c - 1 - (b - \dfrac{2}{3}) = \dfrac{7}{2} - 2 \\
   = \dfrac{{7 - 4}}{2} \\
   = \dfrac{3}{2} \\
  \]
Since the common difference is the same for both consecutive terms we can conclude that
  $a,b - \dfrac{2}{3},c - 1$ are in A.P.
Therefore, option(C) is correct
Now, checking for $\dfrac{c}{a},$
$
  \dfrac{c}{a} = \dfrac{{\left( {\dfrac{9}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} \\
   = 9 \\
  $
Which is an integer
 Therefore, option(D) is correct
Hence, All the options are correct.

Note: We can also verify our answer by putting the value of n in \[\sum\limits_{r = 1}^n {r\left( {r + 1} \right)\left( {2r + 3} \right)} = a{n^4} + b{n^3} + c{n^2} + dn + e\]
Putting \[n = 1\] , we get
\[\sum\limits_{r = 1}^1 {r\left( {r + 1} \right)\left( {2r + 3} \right)} = a{(1)^4} + b{(1)^3} + c{(1)^2} + d(1) + e\]
\[ \Rightarrow (1)\left( {1 + 1} \right)\left( {2(1) + 3} \right) = a + b + c + d + e\]
\[ \Rightarrow a + b + c + d + e = 2(5)\]
\[ \Rightarrow a + b + c + d + e = 10\]
Substituting the values of a, b, c, d, and e
\[a + b + c + d + e = \dfrac{1}{2} + \dfrac{8}{3} + \dfrac{9}{2} + \dfrac{7}{3} + 0\]
\[ = \dfrac{{1 + 9}}{2} + \dfrac{{8 + 7}}{3}\]
\[ = \dfrac{{10}}{2} + \dfrac{{15}}{3}\]
\[ = 5 + 5\]
\[ = 10\]
Therefore, we can conclude the values of A, B, C, D and are correct.