Advertisements
Advertisements
प्रश्न
Expand the following by using binomial theorem.
`(x + 1/x^2)^6`
Advertisements
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`
APPEARS IN
संबंधित प्रश्न
Expand the following by using binomial theorem.
`(x + 1/y)^7`
Find the middle terms in the expansion of
`(3x + x^2/2)^8`
Sum of the binomial coefficients is
Compute 994
Find the coefficient of x4 in the expansion `(1 + x^3)^50 (x^2 + 1/x)^5`
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)n are equal
If n is a positive integer and r is a non-negative integer, prove that the coefficients of xr and xn−r in the expansion of (1 + x)n are equal
If the binomial coefficients of three consecutive terms in the expansion of (a + x)n are in the ratio 1 : 7 : 42, then find n
In the binomial expansion of (1 + x)n, the coefficients of the 5th, 6th and 7th terms are in AP. Find all values of n
Choose the correct alternative:
The value of 2 + 4 + 6 + … + 2n is
