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Expand the expression by using the Pascal's Triangle and the Binomial Theorem.
243a^5+1620a^4b+4320a^3b^2+5760a^2b^3+3840ab^4+1024b^5
l&l&l&l&l&l&l&l&l&l&l&l&l&l Row&&&&&&Pascal's&&&&&&& &&&&&&Triangle&&&&&&& c&c&c&c&c&c&c&c&c&c&c&c 0& & & & & &1 & & & & & 1& & & & &1 & &1 & & & & 2& & & &1 & &2 & &1 & & & 3& & &1 & &3 & &3 & &1 & & 4& &1 & &4 & &6 & &4 & &1 & 5& 1 & & 5 & & 10 & & 10 & & 5 & & 1 Note that each number found in the triangle that is the sum of the two numbers diagonally above it. Now consider the given binomial. ( 3a+ 4b )^5 We can substitute the first term for a and the second term for b using the Binomial Theorem equation and the coefficients from Pascal's Triangle.
| (a+b)^n=P_0a^nb^0+P_1a^(n-1)b^1+... +P_(n-1)a^1b^(n-1)+P_na^0b^n |
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| ( 3a+ 4b)^5= 1( 3a)^5( 4b)^0+ 5( 3a)^4( 4b)^1+ 10( 3a)^3( 4b)^2+ 10( 3a)^2( 4b)^3+ 5( 3a)^1( 4b)^4+ 1( 3a)^0( 4b)^5 |
a^0=1
a^1=a
a * 1=a
(a * b)^m=a^m* b^m
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