Find the 5th term in the expanison of 2x+
1/ 3y)^8
Answers
Answer:
Students trying to do this expansion in their heads tend to mess up the powers. But this isn't the time to worry about that square on the x. I need to start my answer by plugging the terms and power into the Theorem. The first term in the binomial is "x2", the second term in "3", and the power n is 6, so, counting from 0 to 6, the Binomial Theorem gives me:
(x2 + 3)6 = 6C0 (x2)6(3)0 + 6C1 (x2)5(3)1 + 6C2 (x2)4(3)2 + 6C3 (x2)3(3)3
+ 6C4 (x2)2(3)4 + 6C5 (x2)1(3)5 + 6C6 (x2)0(3)6
Then simplifying gives me
(1)(x12)(1) + (6)(x10)(3) + (15)(x8)(9) + (20)(x6)(27)
+ (15)(x4)(81) + (6)(x2)(243) + (1)(1)(729)
= x12 + 18x10 + 135x8 + 540x6 + 1215x4 + 1458x2 + 729
Expand (2x – 5y)7
I'll plug "2x", "–5y", and "7" into the Binomial Theorem, counting up from zero to seven to get each term. (I mustn't forget the "minus" sign that goes with the second term in the binomial.)
(2x – 5y)7 = 7C0 (2x)7(–5y)0 + 7C1 (2x)6(–5y)1 + 7C2 (2x)5(–5y)2
+ 7C3 (2x)4(–5y)3 + 7C4 (2x)3(–5y)4 + 7C5 (2x)2(–5y)5
+ 7C6 (2x)1(–5y)6 + 7C7 (2x)0(–5y)7
Then simplifying gives me: Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved
(1)(128x7)(1) + (7)(64x6)(–5y) + (21)(32x5)(25y2) + (35)(16x4)(–125y3)
+ (35)(8x3)(625y4) + (21)(4x2)(–3125y5) + (7)(2x)(15625y6)
+ (1)(1)(–78125y7)
= 128x7 – 2240x6y + 16800x5y2 – 70000x4y3 + 175000x3y4 – 262500x2y5
+ 218750xy6 – 78125y7
You may be asked to find a certain term in an expansion, the idea being that the exercise will be way easy if you've memorized the Theorem, but will be difficult or impossible if you haven't. So memorize the Theorem and get the easy points.
What is the fourth term in the expansion of (3x – 2)10?
I've already expanded this binomial, so let's take a look:
(3x – 2)10 = 10C0 (3x)10–0(–2)0 + 10C1 (3x)10–1(–2)1 + 10C2 (3x)10–2(–2)2
+ 10C3 (3x)10–3(–2)3 + 10C4 (3x)10–4(–2)4 + 10C5 (3x)10–5(–2)5
+ 10C6 (3x)10–6(–2)6 + 10C7 (3x)10–7(–2)7 + 10C8 (3x)10–8(–2)8
+ 10C9 (3x)10–9(–2)9 + 10C10 (3x)10–10(–2)10
So the fourth term is not the one where I've counted up to 4, but the one where I've counted up just to 3. (This is because, just as with Javascript, the counting starts with 0, not 1.)
Note that, in any expansion, there is one more term than the number in the power. For instance:
(x + y)2 = x2 + 2xy + y2 (second power: three terms)
(x + y)3 = x3 + 3x2y + 3xy2 + y3 (third power: four terms)
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 (fourth power: five terms)
The expansion in this exercise, (3x – 2)10, has power of n = 10, so the expansion will have eleven terms, and the terms will count up, not from 1 to 10 or from 1 to 11, but from 0 to 10. This is why the fourth term will not the one where I'm using "4" as my counter, but will be the one where I'm using "3".
10C3 (3x)10–3(–2)3 = (120)(2187)(x7)(–8) = –2099520x7
Step-by-step explanation:
Answer:
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Step-by-step explanation:
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