If `Sigma_(r=1)^(n) r(r+1)(2r +3)=an^4+bn^3+cn^2+dn +e ` then

Options

A. a-b=d-c
B. e=0
C. `a,b-2//3,c-1 " are in " in A.P`
D. `(b+d)//a` is an integer

Explanation

To solve the problem, we need to evaluate the summation \( \Sigma_{r=1}^{n} r(r+1)(2r +3) \) and express it in the form \( an^4 + bn^3 + cn^2 + dn + e \). We will then derive the required relationships among the coefficients \( a, b, c, d, \) and \( e \).

### Step 1: Expand the expression
We start by expanding the expression \( r(r+1)(2r + 3) \):
\[ r(r+1)(2r + 3) = r(r(2r + 3) + (2r + 3)) = r(2r^2 + 3r + 2r + 3) = 2r^3 + 5r^2 + 3r \]

### Step 2: Write the summation
Now we can write the summation:
\[ \Sigma_{r=1}^{n} r(r+1)(2r + 3) = \Sigma_{r=1}^{n} (2r^3 + 5r^2 + 3r) \]
This can be split into three separate summations:
\[ = 2\Sigma_{r=1}^{n} r^3 + 5\Sigma_{r=1}^{n} r^2 + 3\Sigma_{r=1}^{n} r \]

### Step 3: Use known formulas for summations
We will use the formulas for the sums of powers:
1. \( \Sigma_{r=1}^{n} r = \frac{n(n+1)}{2} \)
2. \( \Sigma_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \)
3. \( \Sigma_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2 \)

Substituting these into our expression:
\[ = 2\left( \frac{n(n+1)}{2} \right)^2 + 5\left( \frac{n(n+1)(2n+1)}{6} \right) + 3\left( \frac{n(n+1)}{2} \right) \]

### Step 4: Simplify each term
1. \( 2\left( \frac{n(n+1)}{2} \right)^2 = \frac{n^2(n+1)^2}{2} \)
2. \( 5\left( \frac{n(n+1)(2n+1)}{6} \right) = \frac{5n(n+1)(2n+1)}{6} \)
3. \( 3\left( \frac{n(n+1)}{2} \right) = \frac{3n(n+1)}{2} \)

### Step 5: Combine the terms
Now we combine these terms:
\[ \Sigma_{r=1}^{n} r(r+1)(2r + 3) = \frac{n^2(n+1)^2}{2} + \frac{5n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} \]

### Step 6: Find a common denominator
The common denominator for these fractions is 6:
\[ = \frac{3n^2(n+1)^2}{6} + \frac{5n(n+1)(2n+1)}{6} + \frac{9n(n+1)}{6} \]
Combining these gives:
\[ = \frac{3n^2(n+1)^2 + 5n(n+1)(2n+1) + 9n(n+1)}{6} \]

### Step 7: Expand and simplify
Now we expand and simplify the numerator:
1. \( 3n^2(n^2 + 2n + 1) = 3n^4 + 6n^3 + 3n^2 \)
2. \( 5n(n+1)(2n+1) = 10n^3 + 15n^2 + 5n \)
3. \( 9n(n+1) = 9n^2 + 9n \)

Combining these:
\[ 3n^4 + (6n^3 + 10n^3) + (3n^2 + 15n^2 + 9n^2) + (3n + 5n + 9n) = 3n^4 + 16n^3 + 27n^2 + 17n \]

### Step 8: Final expression
Thus, we have:
\[ \Sigma_{r=1}^{n} r(r+1)(2r + 3) = \frac{1}{6}(3n^4 + 16n^3 + 27n^2 + 17n) \]
This gives us:
\[ a = \frac{1}{2}, b = \frac{8}{3}, c = \frac{9}{2}, d = \frac{7}{3}, e = 0 \]

### Step 9: Prove the relationships
Now we can prove the required relationships:
1. \( a - b = d - c \)
\[ \frac{1}{2} - \frac{8}{3} = \frac{7}{3} - \frac{9}{2} \]
Both sides simplify to \( -\frac{13}{6} \).

2. \( e = 0 \) is already established.

3. \( \frac{b + d}{a} \) is an integer:
\[ \frac{\frac{8}{3} + \frac{7}{3}}{\frac{1}{2}} = \frac{15/3}{1/2} = 10 \]

4. \( a, \frac{b - 2}{3}, \frac{c - 1}{r} \) is in AP:
\[ a = \frac{1}{2}, \frac{b - 2}{3} = \frac{8/3 - 2}{3} = \frac{2}{9}, \frac{c - 1}{r} = \frac{9/2 - 1}{r} = \frac{7/2}{r} \]
Check the common difference.