Let us consider the binomial expansion ` (1 + x)^(n) = sum_(r=0)^(n) a_(r) x^(r)`
where ` a_(4) , a_(5) "and " a_(6) ` are in AP , ( n ` lt ` 10 ). Consider another
binomial expansion of ` A = root (3)(2) + (root(4) (3))^(13n)` , the expansion of A
contains some rational terms ` T_(a1),T_(a2),T_(a3),...,T_(am)`
` (a_(1) lt a_(2) lt a_(3) lt ...lt a_(m))`
The value of ` sum_(i=1)^(n) a_(i)` is

Options

A. 63
B. 127
C. 255
D. 511

Explanation

To solve the problem step-by-step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understanding the Binomial Coefficients in AP
We know that \( a_4, a_5, a_6 \) are in Arithmetic Progression (AP). The coefficients \( a_r \) in the binomial expansion of \( (1 + x)^n \) are given by \( a_r = \binom{n}{r} \).

For \( a_4, a_5, a_6 \) to be in AP, we have:
\[ 2a_5 = a_4 + a_6 \]
Substituting the binomial coefficients:
\[ 2 \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \]

### Step 2: Expressing the Binomial Coefficients
Using the properties of binomial coefficients, we can express:
\[ \binom{n}{4} = \frac{n!}{4!(n-4)!}, \quad \binom{n}{5} = \frac{n!}{5!(n-5)!}, \quad \binom{n}{6} = \frac{n!}{6!(n-6)!} \]

### Step 3: Setting Up the Equation
Substituting these into our equation gives:
\[ 2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} \]
Dividing through by \( n! \) (assuming \( n! \neq 0 \)):
\[ 2 \cdot \frac{1}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \]

### Step 4: Simplifying the Equation
Multiply through by \( 5! \cdot (n-5)! \cdot 4! \cdot (n-4)! \cdot 6! \cdot (n-6)! \) to eliminate the denominators:
\[ 2 \cdot 4! = 6 \cdot (n-4) + 4 \cdot (n-5) \]
This simplifies to:
\[ 48 = 6(n-4) + 4(n-5) \]
Expanding and simplifying:
\[ 48 = 6n - 24 + 4n - 20 \]
\[ 48 = 10n - 44 \]
\[ 10n = 92 \implies n = 9.2 \]
Since \( n \) must be an integer and \( n < 10 \), we check \( n = 7 \) and \( n = 8 \).

### Step 5: Finding Valid \( n \)
Testing \( n = 7 \):
\[ 2 \cdot \binom{7}{5} = \binom{7}{4} + \binom{7}{6} \]
Calculating:
\[ 2 \cdot 21 = 35 + 7 \implies 42 = 42 \quad \text{(True)} \]
Testing \( n = 8 \):
\[ 2 \cdot \binom{8}{5} = \binom{8}{4} + \binom{8}{6} \]
Calculating:
\[ 2 \cdot 56 = 70 + 28 \implies 112 = 98 \quad \text{(False)} \]
Thus, \( n = 7 \) is valid.

### Step 6: Finding the Rational Terms in \( A \)
Given \( A = \left(2^{1/3} + 3^{1/4}\right)^{13n} \):
Substituting \( n = 7 \):
\[ A = \left(2^{1/3} + 3^{1/4}\right)^{91} \]

### Step 7: Finding Rational Terms
The rational terms occur when \( r/4 \) is an integer. Therefore, \( r \) must be a multiple of 4. The possible values of \( r \) are \( 0, 4, 8, \ldots, 88 \) (up to 91).

### Step 8: Summation of Coefficients
The sum of coefficients \( \sum_{i=1}^{n} a_i \) where \( n = 7 \):
The total sum of coefficients in the expansion of \( (1 + x)^7 \) is \( 2^7 = 128 \). Since we need to exclude \( a_0 \):
\[ \sum_{i=1}^{7} a_i = 128 - 1 = 127 \]

### Final Answer
Thus, the value of \( \sum_{i=1}^{n} a_i \) is:
\[ \boxed{127} \]