Advertisements
Advertisements
प्रश्न
Expand `(2x^2 -3sqrt(1 - x^2))^4 + (2x^2 + 3sqrt(1 - x^2))^4`
उत्तर
Taking 2x2 as a and `3sqrt(1 - x^2)` as b
We have (a – b)4 + (a + b)4
Now (a – b)4 = 4C0 a3(– b) + 4C2(a2)(– b)2 + 4C3(a)(– b)3 + 4C4(– b)4
4C0 = 1 = 4C4 ; 4C1 = 4 = 4C3 ; 4C2 = `(4 xx 3)/(2 xx 1)` = 6
= a4 – 4a3b + 6a2b2 – 4ab3 + b4
Similarly (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
∴ (a – b)4 + (a + b)4 = 2[a4 + 6a2b2 + b4]
Substituting the value of a and b we get
`2[(2x^2)^4 + 6(2x^2)^2 (3sqrt(1 - x^2))^2 + (3 sqrt(1 - x^2))^4]`
= 2[16x8 + 6(4x4)(9(1 – x2)) + 81(1 – x2)2]
= 2[16x8 + 216x4(1 – x2) + 81(1 – x2)2]
= 2[16x8 + 216x4 – 216x6 + 81 + 81x4 – 162x2]
= 2[16x8 – 216x6 + 297x4 – 162x2 + 81]
= 32x8 – 432x6 + 594x4 – 324x2 + 162
APPEARS IN
संबंधित प्रश्न
Evaluate the following using binomial theorem:
(999)5
Expand the following by using binomial theorem.
`(x + 1/y)^7`
Expand the following by using binomial theorem.
`(x + 1/x^2)^6`
Find the middle terms in the expansion of
`(x + 1/x)^11`
Find the middle terms in the expansion of
`(3x + x^2/2)^8`
Find the term independent of x in the expansion of
`(x - 2/x^2)^15`
Find the term independent of x in the expansion of
`(2x^2 + 1/x)^12`
Show that the middle term in the expansion of is (1 + x)2n is `(1*3*5...(2n - 1)2^nx^n)/(n!)`
The last term in the expansion of (3 + √2 )8 is:
Sum of binomial coefficient in a particular expansion is 256, then number of terms in the expansion is:
Compute 1024
Compute 994
Find the coefficient of x4 in the expansion `(1 + x^3)^50 (x^2 + 1/x)^5`
Find the last two digits of the number 3600
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