How do you expand ${(x + 2)}^{5}$ $(x + 2)^5$ using Pascal’s Triangle?

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How do you expand ${(x + 2)}^{5}$ $(x + 2)^5$ using Pascal’s Triangle?

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Answer 1 · 2015-12-31T07:07:18

$x^{5} + 10 x^{4} + 40 x^{3} + 80 x^{2} + 80 x + 32$ $x^5+10x^4+40x^3+80x^2+80x+32$

Explanation:

The $5 th$ $5"th"$ row of Pascal's triangle is

$1, 5, 10, 10, 5, 1$ $1,5,10,10,5,1$

These values are the coefficients in a binomial expansion to the $5 th$ $5"th"$ power.

${(a + b)}^{5} = a^{5} + 5 a^{4} b + 10 a^{3} b^{2} + 10 a^{2} b^{3} + 5 a b^{4} + b^{5}$ $(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$

Notice the pattern of the exponents: the exponent of $a$ $a$ starts at $5$ $5$ and goes to $0$ $0$ , and $b$ $b$ starts at $0$ $0$ and increases to $5$ $5$ .

Apply the rule to $(x + 2)$ $(x+2)$ :

${(x + 2)}^{5} = x^{5} + 5 x^{4} (2) + 10 x^{3} (2^{2}) + 10 x^{2} (2^{3}) + 5 x (2^{4}) + 2^{5}$ $(x+2)^5=x^5+5x^4(2)+10x^3(2^2)+10x^2(2^3)+5x(2^4)+2^5$

$\Rightarrow x^{5} + 10 x^{4} + 40 x^{3} + 80 x^{2} + 80 x + 32$ $=>x^5+10x^4+40x^3+80x^2+80x+32$

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How do you expand ${(x + 2)}^{5}$ $(x + 2)^5$ using Pascal’s Triangle?

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