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3 years ago

What is the fifth term of (2x + y)^8?

Mathematics

1 answer:

astra-53 [7]3 years ago

8 0

Answer:

Expansion for there will be 9 terms.

(a+b) ^n =nco a^n b^o+ nc1 a^(n-1) b^1+nc2 a^(n-2) b^2+nc3 a(n-3) b^3+nc4 a^(n-4) b^4+...........................ncn a^o b^n.

5th term of (2x+y)^8 = ?

5th term is nC4 a^(n-4) b^4

a=2x, b=y, n=8

therefore 8C4 (2x)^(8-4) (y)^4

8!/((8-4)! 4!) (2x)^4 y^4

40320/576 (16 x^4 ) (y^4)

70 (16x^4)(y^4)

1120 X^4 Y^4

Step-by-step explanation:

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Harlamova29_29 [7]

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$\displaystyle\\\frac{3d-2}{8} =-d+16\frac{1}{4} \\\\\frac{3d-2}{8} =-d+\frac{16*4+1}{4} \\\\\frac{3d-2}{8} =-d+\frac{64+1}{4} \\\\\frac{3d-2}{8} =-d+\frac{65}{4}$

Multiply both parts of the equation by 8:

$\displaystyle\\3d-2=(-d+\frac{65}{4} )(8)\\\\3d-2=-8d+65*2\\\\3d-2=-8d+130\\\\3d-2+2=-8d+130+2\\\\3d=-8d+132\\\\3d+8d=-8d+132+8d\\\\11d=132\\$

Divide both parts of the equation by 11:

$d=12$

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