BINOMIAL EXPANSION 2

SUBJECT: FURTHER MATHEMATICS

CLASS: SSS 1

WEEK SIX

TOPIC: BINOMIAL EXPANSION 2

SUB-TOPICS:

1. Discovering n^th time period.
2. Utility of binomial enlargement

SUB-TOPIC 1

Discovering the nth time period

The binomial enlargement of is given as:

=

This enlargement is true for any pure quantity worth of n, giant or small, however when n is a big pure quantity then is small. Certainly, the bigger the worth of n, the nearer turns into to zero. The n notation for this,

the restrict of as n→ is 0.

Additionally, as n→, the nearer the enlargement above turns into to the sum of phrases:

Right here the ellipsis (…) on the finish of the enlargement implies that the enlargement by no means ends, that’s it has infinite variety of phrases.

Now we will use the sigma notation and write:

Discover the image for infinity (∞) on the high of the sigma, this denotes the truth that the sum is a sum of an infinite variety of phrases.

Examples:

1. What number of phrases will there be within the enlargement of. Discover the fourth time period.

Resolution:

n=11, x=2x and

1. Variety of phrases = n+1 = 11+1 =12
2. From the components,

11-3r = 2

-3r =2-11

-3r = -9; 3r = 9 ; therefore r=3 (i.e. the 4^th time period)

1. Discover the time period x² after which time period unbiased of x within the enlargement of discover the fifth time period:

Resolution:

1. The overall time period is

When it’s x², then 12-2r = 2

r=5 (i.e. the sixth time period)

Therefore,

=

=

=

1. If the time period is to be unbiased of x, then 12-2r = 0

r =6 (i.e. the 7^th time period)

= = =

The unbiased time period of x is

Class exercise

1. Discover the time period unbiased of x within the enlargement
2. What number of phrases will there be within the enlargement?

SUB-TOPIC 2

Utility of Binomial enlargement in approximation

If x may be very small, we will take 1 + nx as an approximation of is taken as 1-nx.

Examples:

Discover the linear approximation of the next:

1. (1.01)⁵ (b) (1.02)⁴ (c) (1.05)³ (d) (0.98)⁵

Resolution: (a) (1.01)⁵

Growth of (1+x)ⁿ = 1+nx= (1+0.01)⁵=1+5×0.01= 1+0.050= 1.050

1. (1.02)⁴

Resolution: Growth of (1+x)ⁿ = 1+nx

(1.02)⁴ = (1+0.02)⁴≈1+4×0.02

= 1+0.080= 1.080

(c) (1.05)^3; Growth of (1+x)ⁿ = 1+nx

(1.05)³ = (1+0.05)⁴≈1+3×0.05= 1+0.15= 1.150

(d) (0.98)⁵

Resolution: Growth of 1+xⁿ = 1+nx

(0.98)⁵ = (1-0.02)⁵≈1-5×0.02= 1-0.1= 0.09

EVALUATION

Discover the linear approximation of the next:

1. (1.002)¹⁵ (ii) (0.99)⁸ (iii) (2.004)⁷

PRACTICE QUESTIONS

1. What number of phrases will there be within the enlargement of
2. Discover the phrases within the enlargement of . Discover the fift time period.
3. Increase and therefore consider (0.98)⁷ right to 3 decimal locations.
4. Discover the linear approximation of the next (a) (2.004)⁷ (b) (1.003)¹¹ (c) (d) .
5. Utilizing the binomial theorem, write down and simplify the primary seven phrases of the enlargement of (1+2x)¹⁰ in ascending powers of x. use your enlargement to point out that 1.2¹⁰˃6.19.

WEEKEND ACTIVITY:

1. Discover the variety of phrases within the enlargement of . Discover the third time period.
2. Discover the time period x² and unbiased of x within the enlargement .
3. Discover the coefficient of x^-4 within the enlargement.
4. Decide the coefficient of from the enlargement of .