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प्रश्न
Answer the following
Find the constant term in the expansion of `(2x^2 - 1/x)^12`
उत्तर
Here a = 2x2, b = `(-1)/x`, n = 12
We have tr+1 = nCr an–r .br
= `""^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.
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