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3 years ago

Write the 5th term in the expansion of (5x-y/2)^7

Mathematics

1 answer:

Nezavi [6.7K]3 years ago

5 0

Answer:

65625/4(x^5)(y²)

Step-by-step explanation:

Using binomial expansion

Formula: (n k) (a^k)(b ^(n-k))

Where (n k) represents n combination of k (nCk)

From the question k = 5 (i.e. 5th term)

n = 7 (power of expression)

a = 5x

b = -y/2

....................

Solving nCk

n = 7

k = 5

nCk = 7C5

= 7!/(5!2!) ------ Expand Expression

=7 * 6 * 5! /(5! * 2*1)

= 7*6/2

= 21 ------

.........................

Solving (a^k) (b^(n-k))

a = 5x

b = -y/2

k = 5

n = 7

Substituting these values in the expression

(5x)^5 * (-y/2)^(7-5)

= (3125x^5) * (-y/2)²

= 3125x^5 * y²/4

= (3125x^5)(y²)/4

------------------------------------

Multiplying the two expression above

21 * (3125x^5)(y²)/4

= 65625/4(x^5)(y²)

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