THE BINOMIAL SERIES 61A
i.e. (0.97)^6 =0.8330, correct to 4 significant
figuresProblem 9. Determine the value of (3.039)^4 ,
correct to 6 significant figures using the binomial
theorem.(3.039)^4 may be written in the form (1+x)nas:(3.039)^4 =(3+ 0 .039)^4=[
3(
1 +0. 039
3)] 4= 34 (1+ 0 .013)^4(1+ 0 .013)^4 = 1 +4(0.013)+(4)(3)
(2)(1)(0.013)^2+(4)(3)(2)
(3)(2)(1)(0.013)^3 +···= 1 + 0. 052 + 0. 001014+ 0. 000008788 +···= 1. 0530228
correct to 8 significant figuresHence (3.039)^4 = 34 (1.0530228)
=85.2948, correct to
6 significant figures
Now try the following exercise.Exercise 33 Further problems on the bino-
mial series- Use the binomial theorem to expand
(a+ 2 x)^4.
[
a^4 + 8 a^3 x+ 24 a^2 x^2- 32 ax^3 + 16 x^4
]- Use the binomial theorem to expand
(2−x)^6.
[
64 − 192 x+ 240 x^2 − 160 x^3- 60 x^4 − 12 x^5 +x^6
]- Expand (2x− 3 y)^4
[
16 x^4 − 96 x^3 y+ 216 x^2 y^2
− 216 xy^3 + 81 y^4
]- Determine the expansion of
(
2 x+2
x) 5
.⎡⎢
⎣32 x^5 + 160 x^3 + 320 x+320
x+160
x^3+32
x^5⎤⎥
⎦- Expand (p+ 2 q)^11 as far as the fifth term.
[
p^11 + 22 p^10 q+ 220 p^9 q^2
+ 1320 p^8 q^3 + 5280 p^7 q^4]- Determine the sixth term of
(
3 p+q
3) 13
.[34749p^8 q^5 ]- Determine the middle term of (2a− 5 b)^8.
[700000a^4 b^4 ] - Use the binomial theorem to determine,
correct to 4 decimal places:
(a) (1.003)^8 (b) (1.042)^7
[(a) 1.0243 (b) 1.3337] - Use the binomial theorem to determine,
correct to 5 significant figures:
(a) (0.98)^7 (b) (2.01)^9
[(a) 0.86813 (b) 535.51] - Evaluate (4.044)^6 correct to 3 decimal
places.
[4373.880]
7.4 Further worked problems on the
binomial seriesProblem 10.(a) Expand1
(1+ 2 x)^3in ascending powersofxas far as the term inx^3 , using the
binomial series.
(b) State the limits ofxfor which the expan-
sion is valid.(a) Using the binomial expansion of (1+x)n, where
n=−3 andxis replaced by 2xgives:
1
(1+ 2 x)^3=(1+ 2 x)−^3