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Sum of odd- or even-indexed coefficients of a trinomial power (put x = 1 and x = -1)

This type of question, asked 3 times: 2021 Paper 5, 2021 Paper 9, 3 Apr 2025. All 3 questions below, oldest first.

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  1. Question 1

    2021 Paper 5

    76
    Let [x]\lbrack x\rbrack denote greatest integer less than or equal to xx. If for n∈N,(1−x+x3)n=∑j=03najxjn\in\mathbb{N},\left( 1 -x+x^{3} \right)^{n}=\sum_{j= 0}^{3n} a_{j}x^{j}, then ∑j=0[3n2]a2j+4∑j=0[3n−12]a2j+1\sum_{j = 0}^{\left\lbrack \frac{3n}{2} \right\rbrack} a_{2j}+ 4\sum_{j = 0}^{\left\lbrack \frac{3n - 1}{2} \right\rbrack} a_{2j}+ 1 is equal to:

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  2. Question 2

    2021 Paper 9

    63
    Let (1+x+2x2)20=a0+a1x+a2x2+…+a40x40\left( 1 +x+ 2x^{2} \right)^{20}=a_{0}+a_{1}x+a_{2}x^{2}+ \ldots +a_{40}x^{40}. then a1+a3+a5+…+a37a_{1}+a_{3}+a_{5}+ \ldots +a_{37} is equal to

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  3. Question 3

    3 Apr 2025

    147
    Let (1+x+x2)10=a0+a1x+a2x2+….+a20x20\left( 1 + x + x^{2} \right)^{10} = a_{0} + a_{1}x + a_{2}x^{2} + \ldots. + a_{20}x^{20}. If (a1+a3+a5+…+a19)−11a2=\left( a_{1} + a_{3} + a_{5} + \ldots + a_{19} \right) - 11a_{2} = 121k121k, then k is equal to .