Question

The terms containing \( x^r y^s \) (for certain r and s) are present in both the expansions of \( (x+y^2)^{13 \) and \( (x^2+y)^{14} \). If \( \alpha \) is the number of such terms, then the sum \( \sum_{r,s} \alpha (r+s) = \) (Note: The sum is over the common terms)}

• A. \( 27 \)
• B. \( 40 \)
• C. \( 18 \)
• D. \( 35 \) Correct Answer

Hint:

For common terms in two binomial expansions \( (a_1+b_1)^{n_1} \) and \( (a_2+b_2)^{n_2} \), find the general term for each. Let the variables be \(x\) and \(y\). Equate the powers of \(x\) from both general terms and equate the powers of \(y\) from both general terms. This will give a system of equations for the indices \(k_1\) and \(k_2\) of the general terms. Solve for integer solutions of \(k_1, k_2\) within their valid ranges.

Answer 1

The Correct Option is C

Step 1: Find the general term for the expansion of \( (x+y^2)^{13} \).
The general term \( T_{k_1+1} \) is given by \( {}^{13}C_{k_1} x^{13-k_1} (y^2)^{k_1} = {}^{13}C_{k_1} x^{13-k_1} y^{2k_1} \).
Here, \( 0 \le k_1 \le 13 \).
The powers are \( r_1 = 13-k_1 \) and \( s_1 = 2k_1 \).

Step 2: Find the general term for the expansion of \( (x^2+y)^{14} \).
The general term \( T_{k_2+1} \) is given by \( {}^{14}C_{k_2} (x^2)^{14-k_2} y^{k_2} = {}^{14}C_{k_2} x^{2(14-k_2)} y^{k_2} = {}^{14}C_{k_2} x^{28-2k_2} y^{k_2} \).
Here, \( 0 \le k_2 \le 14 \).
The powers are \( r_2 = 28-2k_2 \) and \( s_2 = k_2 \).

Step 3: For common terms, the powers of \(x\) and \(y\) must be equal.
Equating powers of \(x\): \( r_1 = r_2 \Rightarrow 13-k_1 = 28-2k_2 \) (Equation A) Equating powers of \(y\): \( s_1 = s_2 \Rightarrow 2k_1 = k_2 \) (Equation B)
Step 4: Solve the system of equations for \(k_1\) and \(k_2\).
Substitute Equation B into Equation A: \[ 13-k_1 = 28-2(2k_1) \] \[ 13-k_1 = 28-4k_1 \] \[ 4k_1 - k_1 = 28-13 \] \[ 3k_1 = 15 \] \[ k_1 = 5 \] Now find \(k_2\) using Equation B: \[ k_2 = 2k_1 = 2(5) = 10 \]
Step 5: Verify that \(k_1\) and \(k_2\) are within their valid ranges.
For \(k_1=5\): \(0 \le 5 \le 13 \), which is valid.
For \(k_2=10\): \(0 \le 10 \le 14 \), which is valid.
Since there is a unique solution \((k_1, k_2) = (5, 10)\), there is exactly one common term.
So, the number of such terms, \( \alpha = 1 \).

Step 6: Find the powers \(r\) and \(s\) for this common term.
Using \(k_1=5\): \( r = 13-k_1 = 13-5 = 8 \) and \( s = 2k_1 = 2(5) = 10 \).
So the common term has powers \(x^8 y^{10}\).
(Check with \(k_2=10\): \( r = 28-2k_2 = 28-2(10) = 8 \) and \( s = k_2 = 10 \).
This matches.
) For this common term, \( r+s = 8+10 = 18 \).

Step 7: Calculate the required sum.
The question asks for "the sum \( \sum_{r,s} \alpha (r+s) \)", where \( \alpha \) is the number of common terms and the sum is over the common terms.
Since \( \alpha=1 \) and there is only one common term with \( (r,s) = (8,10) \), the sum is: \[ \alpha (r+s) = 1 \times (8+10) = 1 \times 18 = 18 \] This matches option (3).