Maharashtra State BoardHSC Science (Electronics) 11th
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Answer the following Find the constant term in the expansion of (2x2-1x)12 - Mathematics and Statistics

Sum

Answer the following

Find the constant term in the expansion of `(2x^2 - 1/x)^12`

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Solution

Here a = 2x2, b = `(-1)/x`, n = 12

We have tr+1 = nCr an–r .b

= `""^12"C"_"r" (2x^2)^(12 - "r") ((-1)/x)^"r"`

= 12Cr (2)12–r .(–1)r .x24–2r .x–r

= 12Cr (2)12–r (–1)r x24–3r 

To get the constant term, we must have

x24–3r = x0

∴ 24 – 3r = 0

∴ r = 8

∴ Constant term = 12C8(2)4(– 1)8

= `(12!)/(8!4!) xx 16 xx 1`

= `(12 xx 11 xx 10 xx 9 xx 8!)/(4 xx 3 xx 2 xx 1 xx 8!) xx 16`

= 7920

∴ Constant term is 7920.

Concept: General Term in Expansion of (a + b)n
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4 | Q II. (10) (ii) | Page 86
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