Let . If , then k is equal to __________. [2025]

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Solution

Q.
Let ${(1 + x + x^{2})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{20} x^{20}$ . If $(a_{1} + a_{3} + a_{5} + . . . + a_{19}) - 11 a_{2} = 121 k$ , then k is equal to __________. [2025]

Ans.
(239)

${(1 + x + x^{2})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{20} x^{20}$

Put x = 1, we have

$3^{10} = a_{0} + a_{1} + a_{2} + . . . . . + a_{20}$ ... (i)

Put x = –1, we have

$1 = a_{0} - a_{1} + a_{2} + . . . . . + a_{20}$ ... (ii)

Subtracting equation (ii) from (i),

$a_{1} + a_{3} + . . . + a_{19} = \frac{3^{10} - 1}{2} = 29524$

Also, ${1 + x (1 + x)}^{10} = 1 + {}^{10}C_{1} x (1 + x) + {}^{10}C_{2} x^{2} {(1 + x)}^{2} + . . .$

$∴ a_{2} = {}^{10}C_{1} + {}^{10}C_{2} = 55$

$∴ k = \frac{(a_{1} + a_{3} + . . . + a_{19}) - 11 a_{2}}{121} = \frac{29524 - 605}{121} = 239$ .

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