If ²⁰C_r is the co-efficient of x^r in the expansion of (1 + x)²⁰, then the value of ${\displaystyle \sum _{r=0}^{20}{r}^2}$ ²⁰C_r is equal to

  $T_{r+1}{=}^{60}{C}_r{\left(x^{\frac{1}{2}}\right)}^{60-r}{\left(x^{-\frac{1}{3}}\right)}^r{\left(5^{\frac{-1}{4}}\right)}^{60-r}{\left(5^{\frac{1}{2}}\right)}^r$

for

$x^{10}\frac{60-r}{2}-\frac{r}{3}=10$    

$\Rightarrow 180-3r-2r=60$             

->r = 24

k = 3 + exponent of 5 in 

= $3+\left(\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]-\left[\frac{24}{5}\right]-\left[\frac{24}{5^2}\right]-\left[\frac{36}{5}\right]-\left[\frac{36}{5^2}\right]\right)$  

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5

15. ${\left[3x^2-2ax+3a^2\right]}^3$

= ${\left[3x^2+a\left(-2x+3a\right)\right]}^3$

We know that by binomial theorem,

${(a+b)}^3$ = $a^3+b^3+3ab\left(a+b\right)$

= $a^3+b^3+3a^{2}b+3a{b}^2$

Then,

${\left[3x^2+a\left(-2x+3a\right)\right]}^3$

= (3x²)³ + ${\left[a\left(-2x+3a\right)\right]}^3$ + $\left[3(3x^2){}^2 a\left(-2x+3a\right)\right]$ + $\left[ 3(3x^2){\left\{a\left(-2x+3a\right)\right\}}^2\right]$

= 27x⁶ + $\left[a^3{\left(-2x+3a\right)}^3\right]$ + $\left[3(9x^4)\left(-2ax+3a^2\right)\right]$ + $\left[3(3x^2)\left\{a^2{\left(3a-2x\right)}^2\right\}\right]$

= 27x⁶ + $\left[a^3\left\{-8x^3+27a^3+3\left(4x^2\right)\left(3a\right)+3\left(-2x\right)\left(9a^2\right)\right\}\right]$ + $\left[-54a{x}^5+81a^2{x}^4\right]$ + [ $\left(9a^2{x}^2\right)$ $\left(9a^2+4x^2-12ax\right)$ ]

= 27x⁶ + [ $-8a^3{x}^3+27a^6+36a^4{x}^2-54a^{5}x$ ] + [ $-54a{x}^5+81a^2{x}^4$ ] + [ $(81a^4{x}^2+36a^2{x}^4-108a^3{x}^3$ ]

= 27x⁶ $- 8a^3{x}^3+27a^6+36a^4{x}^2-54a^{5}x$ $- 54a{x}^5+81a^2{x}^4$ + $81a^4{x}^2+36a^2{x}^4-108a^3{x}^3$

= 27x⁶– 54ax⁵ $+ 117a^2{x}^4$ $- 116a^3{x}^3$ 

14. For (a – b) to be a factor of aⁿ – b ⁿwe need to show (aⁿ – bⁿ) = (a – b)k as k is a natural number.

We have, for positive n

aⁿ = ${(a-b+b)}^n$ = ${\left[\left(a-b\right)+b\right]}^n$

=>aⁿ = ⁿC₀(a – b)ⁿ + ⁿC₁(a – b)^{n -1}b + ⁿC₂(a – b)^{n – 2}b² + ………… +ⁿC_n-1 $\left(a-b\right) b^{n-1}$ + ⁿC_nbⁿ

=>aⁿ= ${\left(a-b\right)}^n$ + ⁿC₁ ${\left(a-b\right)}^{n-1} b$ + ⁿC₂ ${\left(a-b\right)}^{n-2} b^2$ + …………….…+ ⁿC_n-1 $\left(a-b\right) b^{n-1}$ + $b^n$ [Since, ⁿC₀ = 1 and

13. The general term of the expansion ${\left(3+ax\right)}^9$ is

T_r+1 = ⁹C_r $3^{\left(9-r\right)}$ ${\left(ax\right)}^r$

= ⁹C_r $3^{\left(9-r\right)}{a}^r{x}^r$

At r = 2,

T₂₊₁ = ⁹C₂ $3^{\left(9-2\right)}{a}^2{x}^2$

= $\frac{9!}{2!\left(9-2\right)!}$ 3⁷a²x²

= $\mathrm{9\prime }$$\mathrm{8\prime }$$7!$/$\mathrm{2\prime }$$\mathrm{1\prime }$$7!\mathrm{}$ 3⁷a²x²

= 36 ×3⁷a²x²

At r = 3,

T₃₊₁ = ⁹C₃ $3^{\left(9-3\right)}{a}^3{x}^3$

= $\frac{9!}{3!\left(9-3\right)!}$ 3⁶a³x³

= 9'8'7'6!/3'2'1'6! 3⁶a³x³

= 84 ×3⁶a³x³

Given that,

Co-efficient of $x^2$ = co-efficient of $x^3$

=> 36 × $3^7{a}^2$ = 84 ×

1.The general term of the expansion (a + b)ⁿ is given by

T_{r +1} = ⁿC_ra^n–rb^r

So, T₁ = ⁿC₀aⁿ = aⁿ

T₂ = ⁿC₁a^n-1b = $\frac{n!}{1!\left(n-1\right)!}$ a^n-1 b = $\frac{n\times \left(n-1\right)!}{\left(n-1\right)!}$ a^n-1b = na^n-1b

T₃ = ⁿC₂a^n-2b² = $\frac{n!}{2!\left(n-2\right)[!}$ a^n-2b² = $\frac{n \times \left(n-1\right)\times \left(n-2\right)!}{2\times 1\times \left(n-2\right)!}$ a^n-2b² = $\frac{n\left(n-1\right)}{2}$ a^n-2b²

Given,

T₁ = 729

=>aⁿ = 729 ------------------ (1)

T₂ = 7290

=>na^n–1b = 7290 ------------- (2)

T₃ = 30375

=> $\frac{n\left(n-1\right)}{2}$ a^n–2b² = 30375