Question

What are the values of a and b

(x+y)^(7)=x7+7x^(6)y+ax^(5)y^(2)+35x^(4)y^(b)+35x^(3)y^(4)+21x^(2)y^(5)+7xy^(6)+y^(7)

Answer 1

a = 21

b = 3

Explanation:

`\boxed{\begin{minipage}{11cm}\underline{Binomial Series}\\\\$\displaystyle (a+b)^n=a^n+\binom{n}{1}a^(n-1)b^1+\binom{n}{2}a^(n-2)b^2+...+\binom{n}{r}a^(n-r)b^r+...+b^n$\\\\\\Where $\displaystyle \binom{n}{r} \: = \:^(n)\text{C}_(r) = (n!)/(r!(n-r)!)$\\\end{minipage}}`

Given binomial expansion:

`(x+y)^7=x^7+7x^6y+ax^5y^2+35x^4y^b+35x^3y^4+21x^2y^5+7xy^6+y^7`

The coefficients of the given binomial expansion are:

• 1, 7, a, 35, 35, 21, 7, 1

Due to the symmetry of the coefficients in a binomial expansion:

• a = 21

The exponent of the second term y increases by one for each term in the expansion. Therefore:

• b = 3