Answer to Question #346431 in Algebra for bookaddict

Question

a. Expand (𝑎+𝑏)⁵. Hence find the coefficient of 𝑥 in the expansion of (4𝑥+2/9𝑥)⁵

b. The coefficient of 𝑥² in the expansion of (1+𝑥)ⁿ is 45. Given that 𝑛 is a positive integer, find the value of 𝑛.

Answer 1

a.

$(a+b)^5=\dbinom{5}{0}a^5+\dbinom{5}{1}a^4b+\dbinom{5}{2}a^3b^2$

$+\dbinom{5}{3}a^2b^3+\dbinom{5}{4}ab^4++\dbinom{5}{5}b^5$

$=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$

$(4x+\dfrac{2}{9x})^5=1024x^5+\dfrac{5(256)(2)x^4}{9x}$

$+\dfrac{10(64)(4)x^3}{81x^2}+\dfrac{10(16)(8)x^2}{729x^3}+\dfrac{5(4)(16)x}{6561x^4}$

$+\dfrac{32x^4}{59049x^5}=1024x^5+\dfrac{2560x^3}{9}+\dfrac{2560x}{81}$

$+\dfrac{1280}{729x}+\dfrac{320}{6561x^3}+\dfrac{32}{59049x^5}$

b.

$(1+x)^n=\dbinom{n}{0}+\dbinom{n}{1}x+\dbinom{n}{2}x^2+...$

$\dbinom{n}{2}=45$

$\dfrac{n(n-1)}{2}=45$

$n(n-1)=90, n>0$

Then $n=10$