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

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.