The coefficient of t4 in the expansion of 1-t61-t3 is -

The correct option is B 15 (1 t^6)^3(1 t)^ 3 (1 t^18 3t^6+3t^12)(1 t)^ 3 ⇒ coefficient of t^4 in (1 t)^ 3 is #160; ^3+4 ...

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General Term of Binomial Expansion

The coefficie...

Question

The coefficient of $t^{4}$ in the expansion of ${(\frac{1 - t^{6}}{1 - t})}^{3}$ is :

12

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15

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10

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14

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t^{4}

{(\frac{1 - t^{6}}{1 - t})}^{3}

t_{0}, t_{1}, t_{2}, t_{n}

(x + a i)^{n}

(t_{0} - t_{2} + t_{4} - t_{6} + \dots .)^{2} + (t_{1} - t_{3} + t_{5} \dots .)^{2} =

T_{0}, T_{1}, T_{2}, . . . . . T_{n}

(x + a)^{n}

(T_{0} - T_{2} + T_{4} - T_{6} + . . . . .)^{2} + (T_{1} - T_{3} + T_{5} + . . . . .)^{2}

Find the G.P. if

(a) T₃ : T₆ = 1 : 8 and T₅ = 64

(b) T₄ : T8 = 1 : 81 and T₆ = 486

(d) S₈ : S₄ = 97 : 81 and T₅ = 16

T_{0}, T_{1}, T_{2}, . . ., T_{n}

{(x + a)}^{n}

{(T_{0} - T_{2} + T_{4} - T_{6} + . . .)}_{0}^{2} + {(T_{1} - T_{3} + T_{5} + . . .)}_{1}^{2}

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