Advertisements
Advertisements
प्रश्न
Find the 5th term in the expansion of (x – 2y)13.
उत्तर
General term is tr+1 = nCr xn-r ar
(x – 2y)13 = (x + (-2y))13
Here x is x, a is (-2y) and n = 13
5th term = t5 = t4+1 = 13C4 x13-4 (-2y)4
= 13C4 x9 24 y4
`= (13xx12xx11xx10)/(4xx3xx2xx1)`× 2 × 2 × 2 × 2× x9y4
= 13 × 11 × 10 × 8x9y4
= 13 × 880x9y4
= 11440x9y4
APPEARS IN
संबंधित प्रश्न
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`
Show that the middle term in the expansion of is (1 + x)2n is `(1*3*5...(2n - 1)2^nx^n)/(n!)`
The constant term in the expansion of `(x + 2/x)^6` is
The last term in the expansion of (3 + √2 )8 is:
Expand `(2x^2 - 3/x)^3`
Expand `(2x^2 -3sqrt(1 - x^2))^4 + (2x^2 + 3sqrt(1 - x^2))^4`
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 the binomial coefficients of three consecutive terms in the expansion of (a + x)n are in the ratio 1 : 7 : 42, then find n
