The first three terms of the binomial expansion of (2 – ax)" are 64 – 16bx +100bx?.
Find the value of each of the integers n, a and b.
Ans: n=6, a=5, b=60
Answers
Answer:
n = 6, a = 5, b = 60
Step-by-step explanation:
In a binomial function (a + b)ⁿ expression that represents the terms,
(a + b)ⁿ = [tex]\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k[/tex]
By this formula,
1st term = [tex]\binom{n}{0}a^{n-0}b^0[/tex] = aⁿ
2nd term = [tex]\binom{n}{1}a^{n-1}b^1[/tex] = [tex]n.a^{n-1}b^1[/tex]
3rd term = [tex]\binom{n}{2}a^{n-2}b^2[/tex] = [tex]\frac{n(n-1)}{2}.a^{n-2}.b^2[/tex]
For the binomial expansion initial 3 terms of (2 - ax)ⁿ = 64 - 16bx + 100bx²
Terms of (2 - ax)ⁿ = [tex]2^n+n(2)^{n-1}(-ax)+\frac{n(n-1)}{2}(2)^{n-2}(-ax)^2[/tex]
= [tex]2^n-n(2)^{n-1}(ax)+n(n-1)(2)^{n-3}(a^2x^2)[/tex]
Comparing the terms of both the expansions,
1st term
2ⁿ = 64
2ⁿ = 2⁶
n = 6
2d term
[tex]n(2)^{n-1}(ax)=16bx[/tex]
[tex]6(2)^{6-1}(a)=16b[/tex]
192a = 16b
b = 12a -----(1)
3rd term
[tex]n(n-1)2^{n-3}(a^2x^2)=100bx^2[/tex]
[tex]6(6-1)2^{(6-3)}(a^2)=100b[/tex]
[tex]30(2)^3(a^2)=100b[/tex]
240a² = 100b
b = 2.4a² -----(2)
From equation (1) and (2),
b = 12a = 2.4a²
a = [tex]\frac{12}{2.4}=5[/tex]
From equation (1)
b = 12a = 60
Therefore, n = 6, a = 5, b = 60
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Answers
Answer:
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Step-by-step explanation:
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Step-by-step explanation:
1. 133 1/3% of 36=?
2. 260% of 360=?
Answers
Answer:
1. 133 1/3% of 36=48
2. 260% of 360=936
Step-by-step explanation:
1. 133 1/3% of 36
=(133 1/3)% of 36
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---------------------------------------------------------------------------------------
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Answers
Answer:
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Step-by-step explanation:
according to the question
(x+2)(x-2)=96
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Please check the explanation.
Step-by-step explanation:
As we know that the area of a rectangle is defined by multiplying the length by the width.
[tex]A=l\times w[/tex]Given
rectangular frame length = l = (x+2) unitsrectangular frame width = w = (x-2) unitsArea = 96 square unitssubstituting all the given values in the formula to find the value of x.
[tex]A=l\times w[/tex]
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[tex]x^2-4=96[/tex]
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[tex]x^2-100=0[/tex]
[tex]x^2=100[/tex]
[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=\sqrt{100},\:x=-\sqrt{100}[/tex]
[tex]x=10,\:x=-10[/tex]
Putting x = -10 in the length and width will make the length and width negative, which can not be possible.
i.e.
length = l = x+2 = -10+2 = -8 units
width = w = x-2 = -10-2 = -12 units
Therefore, x=-10 must be excluded.
Now, putting the length of x = 10.
i.e.
length = l = 10+2 = 10+2 = 12 units
width = w = x-2 = 10-2 = 8
[tex]A=l\times w[/tex]
96 = 12 × 8
96 = 96
Therefore, the correct value of x = 10
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Answers
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Step-by-step explanation:
Answer:
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Step-by-step explanation:
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