1)aro118xample 2. Find the 13th term in the expansion of 9x-3NCERT

1.	1)aro118xample 2. Find the 13th term in the expansion of 9x-3NCERT
Answer» 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)^rfor 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»

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)^rfor 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

Discussion