Question

Find the coefficient of x^5 in the expansion of (1-x)^5(1+x+x^2+x^3)^4.

Hint: Binomial Theorem

Answer 1

The coefficient of x⁵ in the expansion of (1 - x)⁵ (1 + x + x² + x³)⁴ is 4.

Step-by-step explanation:

Now, (1 - x)⁵ (1 + x + x² + x³)⁴

= (1 - x) (1 - x)⁴ (1 + x + x² + x³)⁴

= (1 - x) {(1 - x) (1 + x + x² + x³)}⁴

= (1 - x) (1 - x⁴)⁴

= (1 - x) (⁴C₀ - ⁴C₁ x⁴ + ⁴C₂ x⁸ - ⁴C₃ x¹² + ⁴C₄ x¹⁶)

= (1 - x) (1 - 4x⁴ + 6x⁸ - 12x¹² + x¹⁶)

= 1 - 4x⁴ + 6x⁸ - 12x¹² + x¹⁶ - x + 4x⁵ - 6x⁹ + 12x¹³ - x¹⁷

∴ the coefficient of x⁵ is 4.

Binomial formulae:

• (x - a)ⁿ = ⁿC₀ xⁿ - ⁿC₁ xⁿ⁻¹ . a + ⁿC₂ xⁿ⁻² . a² - ... + (- 1)ⁿ . ⁿCₙ . aⁿ
• 1 - xⁿ = (1 - x) (1 + ... + xⁿ⁻² + xⁿ⁻¹)

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Answer 2

Answer:

We know that for a Geometric Progression (GP), the sum S of all the terms is given by-

$s = a + ar + ar {}^{2} + ar {}^{n - 1} \\ s = a \frac{r {}^{n} - 1 }{r - 1}$

where a is the first term , r is the common ratio and n is the number of terms of GP.

Now coming back to question ,we need to simplify -

$\frac{1 + x + x {}^{2} + x {}^{3} + x {}^{2018} }{1 + x {}^{3} + x {}^{6} + x {}^{2016} }$

We can clearly see that both the numerator as well as the denominator are in 2 different GP.

So we can simplify both series separately and then compute the final answer.