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2 years ago

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Middle term in expansion of. (x+1/x)^18

1 answer:

ladessa [460]2 years ago

3 0

Answer:

Step-by-step explanation:

I used Pascal's Triangle to the 19th row (that took forever!!) and the leading coefficients are as follows:

1, 18, 153, 816, 3060, 8568, 18564, 31824, 43758, 48620, 43758, 31824, 18564, 8568, 3060, 816, 153, 18, 1

You need the middle term so we can stop the expansion at 48,620. We will choose an a and a b from the binomial to be a = x and b = 1/x. Start with the coefficients from above in order and follow the pattern:

1(x)¹⁸(1/x)⁰ + 18(x)¹⁷(1/x)¹ + 153(x)¹⁶(1/x)² + 816(x)¹⁵(1/x)³ + 3060(x)¹⁴(1/x)⁴ + 8568(x)¹³(1/x)⁵ + 18564(x)¹²(1/x)⁶ + 31824(x)¹¹(1/x)⁷ + 43758(x)¹⁰(1/x)⁸ + 48620(x)⁹(1/x)⁹

That last term there is the middle term. Simplified it looks like this:

$\frac{48620x^9}{x^9}$ which is simplified to just 48620

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Step-by-step explanation:

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2 years ago

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Hatshy [7]

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1 year ago

Let f(x)=3x^2+2.

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2 years ago

The indefinite integral can be found in more than one way. First use the substitution method to find the indefinite integral. Th

Fantom [35]

Answer:

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C$

Step-by-step explanation:

To find:

∫ $6x^5(x^6-2)\,dx$

Solution:

Method of substitution:

Let $x^6-2=t$

Differentiate both sides with respect to $t$

$6x^5\,dx=dt$

[use $(x^n)'=nx^{n-1}$ ]

So,

∫ $6x^5(x^6-2)\,dx$ = ∫ $t\,dt$ = $\frac{t^2}{2}+C_1$ where $C_1$ is a variable.

(Use ∫ $t^n\,dt=\frac{t^{n+1} }{n+1}$ )

Put $t=x^6-2$

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1$

Use $(a-b)^2=a^2+b^2-2ab$

So,

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C$

where $C=2+C_1$

Without using substitution:

∫ $6x^5(x^6-2)\,dx$ = ∫ $6x^{11}-12x^5\,dx$ = $\frac{6x^{12} }{12}-\frac{12x^6}{6}+C=\frac{x^{12} }{2}-2x^6+C$

So, same answer is obtained in both the cases.

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1 year ago

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Lyrx [107]

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Step-by-step explanation:

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2 years ago