Solution for Section 6.7 Question 1
1a) By the Binomial Theorem:
| (x + y)4 |
= |
4 |
( |
4 |
) |
x4-k yk |
|
| S |
| k |
| k=0 |
|
|
|
|
= |
| x4 |
+ |
( |
4 |
) |
x3 y1 |
+ |
( |
4 |
) |
x2 y2 |
+ |
( |
4 |
) |
x1 y3 |
+ |
y4 |
|
|
|
|
| 1 |
2 |
3 |
|
|
|
|
|
= |
| x4 |
+ |
4 x3 y1 |
+ |
6 x2 y2 |
+ |
4 x1 y3 |
+ |
y4. |
|
b) By the Binomial Theorem:
| (2a - 3b)5 |
= |
5 |
( |
5 |
) |
x5-k yk |
|
| S |
| k |
| k=0 |
|
|
|
|
= |
| (2a)5 |
+ |
( |
5 |
) |
(2a)4 (-3b)1 |
+ |
( |
5 |
) |
(2a)3 (-3b)2 |
+ |
( |
5 |
) |
(2a)2 (-3b)3 |
+ |
( |
5 |
) |
(2a)1 (-3b)4 |
+ (-3b)5 |
| 1 |
2 |
3 |
4 |
|
|
|
|
|
= |
32 a5 - 240 a4
b + 720 a3 b2 - 1080 a2 b3
+ 810 a b4 - 243 b5. |
Back to Section 6.7