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Row 1.3cm Pascal's Triangle 1.2cm cccccccccccc 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 greater than 1 found in the triangle is the sum of the two numbers diagonally above it. Now consider the given binomial. ( x+ 3 )^5 We can substitute the first term for a and the second term for b using the Binomial Theorem equation and use the coefficients from the {\color{#FF0000}{5}}^\text{th} row of Pascal's Triangle.
| (a+b)^n= _nC_0a^nb^0+ _nC_1a^(n-1)b^1+... + _nC_(n-1)a^1b^(n-1)+ _nC_na^0b^n |
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| ( x+ 3)^5= 1 x^5( 3)^0+ 5 x^4( 3)^1+ 10 x^3( 3)^2+ 10 x^2( 3)^3+ 5 x^1( 3)^4+ 1 x^0( 3)^5 |
a^0=1
a^1=a
a * 1=a
Calculate power
Multiply
Row 1.3cm Pascal's Triangle 1.2cm cccccccccccc 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 greater than 1 found in the triangle is the sum of the two numbers diagonally above it. Now consider the given binomial. (2p-q)^4 ⇔ ( 2p+( - q) )^4 We can substitute the first term for a and the second term for b using the Binomial Theorem equation and use the coefficients from the {\color{#FF0000}{4}}^\text{th} row of Pascal's Triangle.
| (a+b)^n=_nC_0a^nb^0+ _nC_1a^(n-1)b^1+... + _nC_(n-1)a^1b^(n-1)+ _nC_na^0b^n |
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| ( 2p+( q))^4= 1( 2p)^4( - q)^0+ 4( 2p)^3( - q)^1+ 6( 2p)^2( - q)^2+ 4( 2p)^1( - q)^3+ 1( 2p)^0( - q)^4 |
a^0=1
a^1=a
a * 1=a
(a * b)^m=a^m* b^m
Calculate power
Multiply