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Exercises
- Write down the first four terms in ascending power of x of the binomials expansion of
- The sum of the nth terms of a progression is given by Sn where Sn = n(3n − 4). Show that the progression is an AP.
- The sum of the nth terms of an AP is Sn where Sn = n² − 3n. Write down the fourth term and the nth term.
Notice: To solve these problems, we can use Pascal's Triangle, but that needs a lot of time! An alternative way is to use the Binomial Theorem, which offers a quicker method for expanding any complex binomial series. Only and only when the power n is a positive integer. see (a + b)n
(a) (1 + 3x)12
= 12c0·112·(3x)0 +
12c1·111·(3x)1 +
12c2·110·(3x)2 +
12c3·19·(3x)3 + ... + (3x)12
= 1 + 36x + 594x² + 5940x3 + ... + 531441x12
(b) (1 − 2x)9
= 9c0·19·(−2x)0 +
9c1·18·(−2x)1 +
9c2·17·(−2x)2 +
9c3·16·(−2x)3 + ... + (−2x)9
= 1 − 18x + 144x² − 672x3 + ... − 512x9
(c) (2 − x)10
= 10c0·110·(−x)0 +
10c1·19·(−x)1 +
10c2·18·(−x)2 +
10c3·17·(−x)3 + ... + (−x)10
= 1024 − 5120x + 11520x² − 15360x3 + ... + x10
| = 20c0·120·(−x/3)0 +
20c1·119·(−x/3)1 +
20c2·118·(−x/3)2 +
20c3·117·(−x/3)3 + ... +
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| = 1 − 20x/3 + 190x²/9 − 1140x3/27 + ... + x20/320 |
= 7c0·27·(−3/2x)0 +
7c1·26·(−3/2x)1 +
7c2·25·(−3/2x)2 +
7c3·24·(−3/2x)3 + ... + ...
= 128 − 672/x + 1512/x² − 1890/x3 + ... + ...
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| = 9c0·(3/2)9·(2x)0 +
9c1·(3/2)8·(2x)1 +
9c2·(3/2)7·(2x)2 +
9c3·(3/2)6·(2x)3 + ... +
= (3/2)9 + (310/27)x + (39/8)x² + (7·(37)/2)x3 + ... + ...
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the first nth terms are given by an = Sn − Sn−1 |
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⇒ an = n(3n − 4) − (n − 1) [3(n − 1) − 4]
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= 3n² − 4n − (n − 1) (3n − 7)
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= 3n² − 4n − 3n² + 10n − 7)
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| ⇒ an = 6n − 7, therefore, we can find out: |
a1 = 6·1 − 7 = − 1 ⤸+6
a2 = 6·2 − 7 = 5 ⤸+6
a3 = 6·3 − 7 = 11 ⤸+6
a4 = 6·4 − 7 = 17 ⤸+6
... ... ...
... ... ...
an = 6·n − 7
and the sequence is: − 1, 5, 11, 17, 23,...,
As you can see, the common difference d = 6.
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By replacing a1 and an in the AP's general formula, we should obtain
Sn = n(3n − 4), which is the proof that Sn is an AP progression.
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By replacing a1 and an in the AP's general formula, we should obtain
Sn = n(3n − 4), which is the proof that Sn is an AP progression.
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| Sn = |
n(a1 + an)
2
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= |
n(−1 + 6n − 7)
2
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= |
2n(3n − 4)
2
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| ⇒ Sn = n(3n − 4), the progression is an AP. |
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| * The nth term is given by: an = Sn − Sn−1 |
| ⇒ an = n² − 3n − [(n−1)² − 3(n−1)] |
| = n² − 3n − (n² − 5n + 4) |
| = n² − 3n − n² + 5n − 4 |
⇒ an = 2n − 4
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| * The 4th term a4 = 4 |
- Find the first three terms in the expansion of (1 + b)5.
Use the substitution x = 1 + b, to find the coefficient of b² in the expansion
of (1 + b − 2b²)5.
* To expand (1 + b)5, we use the Binomial Theorem:
∴ (1 + b)5 = 5C0(1)5(6)0 +
5C1(1)4(6)1 +
5C2(1)3(6)2 + ... +
5C5(1)0(6)5
= 1 + 5 + 10
⇒ (1 + b)5 = 1 + 5 + 10, which are the 1st three terms of the expansion.
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* Now x = 1 + b is plugged in the expression (1 + b − 2b²)5 by substitution
⇒ (x − 2b²)5.
We can still use the Binomial Theorem, but let's go for the Pascal's Triangle.
| 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 |
As you can see, we're concerned by the 5th line because our power is 5
∴ we can write:
1(x)5(−2b²)0 +
5(x)4(−2b²)1 +
10(x)3(−2b²)2 +
10(x)2(−2b²)3 +
5(x)1(−2b²)4 +
1(x)0(−2b²)5
and after working out the parenthesis we obtain:
x5 − 10x4b² + 40x3b4 − 80x²b6 + 80xb8 − 32xb10.
Then, the coefficient of b² is − 10
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- The coefficient of b² and b³ in the expansion of (a + b)6 are equal.
Find the value of a.
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