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प्रश्न
Expand the following by using binomial theorem.
`(x + 1/y)^7`
उत्तर
(x + a)n = nC0 xn a0 + nC1 xn-1 a1 + nC2 xn-2 a2 +...+ nCr xn-r ar +...+ nCn an
∴ `(x + 1/y)^7 = 7"C"_0x^7 - 7"C"_1x^6 (1/y)^1 + 7"C"_2x^5 (1/y)^2 + 7"C"_3x^4 (1/y)^3 + 7"C"_4x^3 (1/y)^4 + 7C_5x^2 (1/y)^5 + 7"C"_6x (1/y)^6 + 7"C"_7 (1/y)^7`
`= x^7 + 7x^6 (1/y) + (7xx6)/(2xx1) x^5 (1/y^2) + (7xx6xx5)/(3xx2xx1) x^4 (1/y^3) + (7xx6xx5xx4)/(4xx3xx2xx1) x^3 (1/y^4) + (7xx6xx5xx4xx3)/(5xx4xx3xx2xx1) x^2 (1/y^5) + 7x (1/y^6) + (1/y^7)`
`= x^7 + (7x^6)/y + (21x^5)/y^2 + (35x^4)/y^3 + (35x^3)/y^4 + (21x^2)/y^5 + (7x)/y^6 + 1/y^7`
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