Using Pascal's Triangle to Solve a sqrt Binomial

May 2010
1
0
Gallifrey
I need an answer to these expressions using the Pascal method... I was doing very well on my work until I ran into these two questions:

\(\displaystyle
(\sqrt{a}+\sqrt{b})^6
\)

and

\(\displaystyle
(x + 1/x)^4
\)

If anyone could help me on these, it'd be greatly appreciated. (Bow)
 

e^(i*pi)

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Feb 2009
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West Midlands, England
I need an answer to these expressions using the Pascal method... I was doing very well on my work until I ran into these two questions:

\(\displaystyle
(\sqrt{a}+\sqrt{b})^6
\)

and

\(\displaystyle
(x + 1/x)^4
\)

If anyone could help me on these, it'd be greatly appreciated. (Bow)
Pascal's triangle only works for the coefficients of each term and I've never heard of Pascal's method. The easiest way would be the binomial theorem

\(\displaystyle (a+b)^n = \sum_{k=0}^n {n \choose k}a^{n-k}b^k\)

Where \(\displaystyle {n \choose k} = \frac{n!}{(n-k)!k!}\)


Taking your first example (n=6) I will take the first two terms

\(\displaystyle (a^{\frac{1}{2}} + b^{\frac{1}{2}})^6 = {6 \choose 0} (a^{\frac{1}{2}})^3 (b^{\frac{1}{2}})^0 + {6 \choose 1} (a^{\frac{1}{2}})^{6-1} (b^{\frac{1}{2}})^1\)

Simplifying that we get \(\displaystyle a^3 + 6a^{\frac{5}{2}}b^{\frac{1}{2}}\)


Have a go at the rest of it and post if you don't understand
 
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