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प्रश्न
Expand the following by using binomial theorem.
`(x + 1/x^2)^6`
उत्तर
(x + a)n = nC0 xn a0 + nC1 xn-1 a1 + nC2 xn-2 a2 +...+ nCr xn-r ar +...+ nCn an
∴ `(x + 1/x^2)^6 = 6"C"_0x^6 + 6"C"_1x^5 (1/x^2)^1 + 6"C"_2x^4 (1/x^2)^2 + 6"C"_3x^3 (1/x^2)^3 + 6"C"_4x^2 (1/x^2)^4 + 6"C"_5x(1/x^2)^5 + 6"C"_6(1/x^2)^6`
`= x^6 + 6x^3 + (6xx5)/(2xx1) x^4 1/x^4 + (6xx5xx4)/(3xx2xx1) x^3 (1/x^6) + (6xx5)/(2xx1) x^2 (1/x^8) + 6x(1/x^10) + 1(1/x^12)`
`= x^6 + 6x^3 + 15 + 20 1/x^3 + 15 1/x^6 + 6(1/x^9) + 1/x^12`
`= x^6 + 6x^3 + 15 + 20/x^3 + 15/x^6 + 6/x^9 + 1/x^12`
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