Question

Can anyone help me solve this question?

e. Use the expansion of (1 + x)^5 to find the exact value of (1 + √5)^5

Answer 1

just a quick addition to "sqdancefan" superb reply above

`\qquad \qquad \textit{binomial theorem expansion} \\\\ \qquad \qquad (1+√(5))^(5)~\hspace{4em} \begin{array}{clcl} term&coefficient&value\\ \cline{1-3}&\\ 1&+1&(1)^(5 )(√(5))^0\\ 2&+5&(1)^(4)(√(5))^1\\ 3&+10&(1)^(3)(√(5))^2\\ 4&+10&(1)^(2)(√(5))^3\\ 5&+5&(1)^(1)(√(5))^4\\ 6&+1&(1)^(0)(√(5))^5 \end{array} \\\\[-0.35em] ~\dotfill`

`1(1)+5(1)(√(5))^1+10(√(5))^2+10(1)(√(5))^3+5(√(5))^4+(√(5))^5 \\\\\\ 1+5√(5)+50+50√(5)+125+25√(5)\implies {\Large \begin{array}{llll} 176+80√(5) \end{array}}`

Answer 2

176 +80√5

Explanation:

You want the exact value of (1 +√5)^5.

Binomial expansion

The coefficient of the expansion are found in Pascal's triangle (attached). Row 5 is used for the 5th power:

(a +b)^5 = a^5 +5·a^4·b + 10·a^3·b^2 +10·a^2·b^3 +5·a·b^4 +b^5

Application

When a = 1 and b = √5, this becomes ...

(1 +√5)^5 = 1 + 5·√5 +10·5 + 10·5√5 +5·25 +25√5

= (1 +50 +125) +(5 +50 +25)√5

(1 +√5)^5 = 176 +80√5