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Lera25 [3.4K]
3 years ago
14

The coefficient of x^ky^n-k in the expansion of (x+y)^n equals (nk). True or false.

Mathematics
1 answer:
AVprozaik [17]3 years ago
7 0

Answer:

The correct option is;

False

Step-by-step explanation:

The coefficient of x^k·y^(n-k) is nk,   False

The kth coefficient of the binomial expansion, (x + y)ⁿ is  \dbinom{n}{k} = \dfrac{n!}{k!\cdot (n-k)!} = C(n,k)

Where;

k = r - 1

r = The term in the series

For an example the expansion of (x + y)⁵, we have;

(x + y)⁵ = x⁵ + 5·x⁴·y + 10·x³·y² + 10·x²·y³ + 5·x·y⁴ + y⁵

The third term, (k = 3) coefficient is 10 while n×k = 3×5 = 15

Therefore, the coefficient of x^k·y^(n-k) for the expansion  (x + y)ⁿ = C(n,k) not nk

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Read 2 more answers
I need the correct answer please
nikdorinn [45]
So first step is to simplify everything outside of the radicals.
6*2=12
:. The expression is
__ __
12*\| 8 * \| 2
Now we know that
__ __ __
\| 8 = \| 4 * \| 2

And
__ __
\| 2 * \| 2 = 2

And
__
\| 4 = 2

So if we incorporate what we know into the equation, we can figure it out.
So let's first expand the radical 8.
__ __ __
12*\| 4 * \| 2 * \| 2

Now by simplifying the radical four and combining the radical twos, we can get all whole numbers.
12*2*2
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Answer:48
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3 years ago
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