Let a+bx+cx210=∑i=020 pixi,a,b,c∈N. If p1 = 20, and p2 = 210 then 2(a + b + c) is equal

General term is=10!r1!r2!r3!(a)r1(bx)r2cx2r3 For coeff. of x:r2+2r3=1 r1 r2 r39 1 0 ∴p1= coef. of x=10!9!a9b1=20 ⇒a9⋅b=2 Coff of x2:10!8!2!0!a8⋅b2+10!9!0!1!⋅a9⋅c=210⇒45a8⋅b2+10⋅a9⋅c=210⇒9a8b2+2a9⋅c=42 ∴ a=1,b=2,c=32(a+b+c)=2(3+2+1)=12

Let a+bx+cx210=∑i=020 pixi,a,b,c∈N. If p1 = 20, and p2 = 210 then 2(a + b + c) is equal

General term is=10!r1!r2!r3!(a)r1(bx)r2cx2r3 For coeff. of x:r2+2r3=1 r1 r2 r39 1 0 ∴p1= coef. of x=10!9!a9b1=20 ⇒a9⋅b=2 Coff of x2:10!8!2!0!a8⋅b2+10!9!0!1!⋅a9⋅c=210⇒45a8⋅b2+10⋅a9⋅c=210⇒9a8b2+2a9⋅c=42 ∴ a=1,b=2,c=32(a+b+c)=2(3+2+1)=12

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Let ${(a + bx + {cx}^{2})}^{10} = \sum_{i = 0}^{20} p_{i} x^{i}, a, b, c \in N$ . If p₁ = 20, and p₂ = 210 then 2(a + b + c) is equal

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Detailed Solution

$General term is= \frac{10!}{r_{1}! r_{2}! r_{3}!} (a)^{r_{1}} (b x)^{r_{2}} {(c x^{2})}^{r_{3}} For coeff. of x : r_{2} + 2 r_{3} = 1 \begin{array}{lll} r_{1} r_{2} r_{3} \\ 9 1 0 \end{array} ∴ p_{1} = coef. of x = \frac{10!}{9!} a^{9} b^{1} = 20 \Rightarrow a^{9} \cdot b = 2$
$Coff of x^{2} : \frac{10!}{8! 2! 0!} a^{8} \cdot b^{2} + \frac{10!}{9! 0! 1!} \cdot a^{9} \cdot c = 210$
$\begin{array}{l} \Rightarrow 45 a^{8} \cdot b^{2} + 10 \cdot a^{9} \cdot c = 210 \\ \Rightarrow 9 a^{8} b^{2} + 2 a^{9} \cdot c = 42 \end{array} \begin{aligned} ∴ a = 1, b = 2, c = 3 \\ 2 (a + b + c) = 2 (3 + 2 + 1) = 12 \end{aligned}$