(b)
4
4
x?
7
1
2x?
(S
Find the middle ter n(s) in the expansion of (2x^2-1/x^3)^7
Answers
Answer:
wht??
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Answer:
Step-by-step explanation:
First you need to know the binomial theorem.
What is
- (a-b)²....... IT is 1.a² - 2.ab +1.b²
- (a-b)³........IT is 1.a³ - 3.a²b +3.ab²-b³
DO YOU SEE A PATTERN?
First the sign is alternating....
secondly....The power of a in each term is reducing...and b i increasing.
WHAT ABOUT THE COEFFICIENTS?
ITS PART OF THE Pascal's Triangle.
1 2 1
1 3 3 1
1 4 6 4 1
....
SO THE QUESTION IS About THE 7TH POWER
THE coefficients are
1 7 21 35 35 21 7 1
So (2x²-1/x³)^7 =
128x^14+448x^9+672x^4+560/x+280/x^6+84/x^11+14/x^16+1/x^21
SO 560/x + 280/x^6 is ARE Middle terms of the expansion
IT TOOK ME A WHILE TO CALCULATE BY HAND SO HOPE YOU APPRECIATE IT