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Absolute Value Function

If 3 x-17=a7 ...

Question

If (3x − 1)⁷ = a₇x⁷ + a₆x⁶ + a₅x⁵ +...+ a₁x + a₀, then a₇ + a₅+ ...+a₁ + a₀=

(a) 0

(b) 1

(c) 128

(d) 64

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Solution

Given that,

Putting

We get

Hence, the correct option is (c).

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Similar questions

(3 x - 1)^{7} = a_{7} x^{7} + a_{6} x^{6} + a_{5} x^{5} + . . . + a_{0}

a_{7} + a_{6} + . . . a_{0} = ?

(3 x - 1) = a_{7} x^{7} + a_{6} x^{6} + . . . . + a_{0}

a_{7} + a_{6} + . . . + a_{0}

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

(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

I f (1 + x)^{10} = a_{0} + a_{1} x + . . . . a_{10} x^{10}, t h e n (a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

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

, then value of

(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})^{2} + (a_{1} - a_{3} + a_{5} - a_{7} + a_{9})^{2}

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