NIMCET 2009 — Mathematics PYQ

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Question

NIMCET 2009 — Mathematics PYQ

NIMCET | Mathematics | 2009

If $(1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + \dots + a_{12}x^{12}$ , then the value of $a_2 + a_4 + a_6 + \dots + a_{12}$ is:

Choose the correct answer:

Explanation

1. Given Equation

The given expansion is:

(1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + a_3x^3 + \dots + a_{12}x^{12}

2. Finding the Sum of All Coefficients

To find the sum of all coefficients, substitute $x = 1$ into the equation:

(1 + 1 - 2(1)^2)^6 = 1 + a_1 + a_2 + a_3 + \dots + a_{12}

(2 - 2)^6 = 1 + a_1 + a_2 + a_3 + \dots + a_{12}

0 = 1 + a_1 + a_2 + a_3 + \dots + a_{12} \quad \text{--- (Equation 1)}

3. Finding the Alternating Sum of Coefficients

To find the alternating sum, substitute $x = -1$ into the equation:

(1 + (-1) - 2(-1)^2)^6 = 1 + a_1(-1) + a_2(-1)^2 + a_3(-1)^3 + \dots + a_{12}(-1)^{12}

(1 - 1 - 2)^6 = 1 - a_1 + a_2 - a_3 + \dots + a_{12}

(-2)^6 = 1 - a_1 + a_2 - a_3 + \dots + a_{12}

64 = 1 - a_1 + a_2 - a_3 + \dots + a_{12} \quad \text{--- (Equation 2)}

4. Isolating the Even Terms

Add Equation 1 and Equation 2:

(1 + a_1 + a_2 + \dots + a_{12}) + (1 - a_1 + a_2 - \dots + a_{12}) = 0 + 64

The odd terms ( $a_1, a_3, \dots$ ) cancel out:

2(1 + a_2 + a_4 + a_6 + \dots + a_{12}) = 64

Divide by 2:

1 + a_2 + a_4 + a_6 + \dots + a_{12} = 32

5. Calculate the Final Value

We need the value of $a_2 + a_4 + a_6 + \dots + a_{12}$ :

a_2 + a_4 + a_6 + \dots + a_{12} = 32 - 1

a_2 + a_4 + a_6 + \dots + a_{12} = 31

Correct Option:

(d) 31