Expansion of common powers
Expansion of common powers
(a+b)^0=1 (a+b)^1=a+b (a+b)^2=a^2+2ab+b^2 (a+b)^3=a^3+3a^2b+3ab^2+b^3 (a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b (a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5 (a+b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 - …
Pascal’s triangle
The coefficients in the expansions above correspond to the rows of Pascal’s triangle:
row |
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
6 | 1 6 15 20 15 6 1
^ ^
|_ column 0 |_ column 5
We can write the the coefficient at row
Finding specific terms using a calculator
If we want to find the coefficient of the
Finding a specific term using factorials
If we want to find
Finding all the coefficients
Expand (2+x)^4
(2+x)^4 = {^4C_0 (2)^4 (x)^0} + {^4C_1 (2)^3 (x)^1} + {^4C_2 (2)^2 (x)^2} + {^4C_3 (2)^1 (x)^3} + {^4C_4 (2)^0 (x)^4} = 1 \times 16 \times 1 + 4 \times 8 \times x + 6 \times 4 \times x^2 + 4 \times 2 \times x^3 + 1 \times 1 \times x^4 = 16 + 32x + 24x^2 + 8x^3 + x^4
Expand (1+2x)^6
=1^6
+^6C_1(1)^5(2x)^1
+ ^6C_2(1)^4(2x)^2
+^6C_3(1)^3(2x)^3
+ ^6C_4(1)^2(2x)^4
+ ^6C_5(1)^1(2x)^5
+ ^6C_6(1)^0(2x)^6 =1 + 6 \times 1 \times 2x + 15 \times 1 \times 4x^2 + 20 \times 1 \times 8x^3 + 15 \times 1 \times 16x^4 + 6 \times 1 \times 32x^5 + 1 \times 1 \times 64x^6 =1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6 - Answer:
1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6
Representing the number of combinations
We can write the number of combinations of choosing