Question

Question 6 Find the 13th term in the expansion of [ 9x - 1/3.x^(1/2) ] ^18 , x ≠ 0

Class X1 - Maths -Binomial Theorem Page 171

Answer 1

we know,
General term of the expansion (X - Y)ⁿ is
T_{r+1} = nCr.X^{n-r}(-Y)^r use this here,

here ,
X = 9x
Y = 1/3√x
n = 18

T_{r+1} = 18Cr.(9x)^{18-r}.(1/3√x)^r
for 13th term , we have to use r = 12

now,
T_{12+1} = 18C12.(9x)^{18-12}(1/3√x)^12

= {18!/12!×6!} (9x)^6(1/3√x)^12

= { 18 × 17 × 16 × 15 × 14 × 13/6×5×4×3×2}(9)^6(1/3)^12 { x^6 × 1/√x^12}

= { 17 × 4 × 3 × 7 × 13} (9^6/9^6){x^6 × 1/x^6}

= 17 × 12 × 91

= 18564

hence, 13th term = 18564

Answer 2

General term of expansion(a+b)ň is

Tr+1=ňCraň-rbr

Step-by-step explanation:

we need to calculate 13th term

• putting r=12 n=18a=9x and b =1/3
• $\sqrt{x}$
• put this value in formula