Let ( a + b x + c x 2 ) 10 = i = 0 20 p i x i , a , b , c . If p 1 = 20 and p 2 = 210 , then 2 ( a + b + c ) is equal to [2023]

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Solution

Q.
Let ${(a + b x + c x^{2})}^{10} = \sum_{i = 0}^{20} p_{i} x^{i}, a, b, c \in ℕ$ . If $p_{1} = 20$ and $p_{2} = 210,$ then $2 (a + b + c)$ is equal to [2023]

1 8

2 6

3 12

4 15

Ans.
(3)

We have, ${(a + b x + c x^{2})}^{10} = \sum_{i = 0}^{20} p_{i} x_{i}, a, b, c \in ℕ$ , $p_{1} = 20$ and $p_{2} = 210$

Now ${(a + b x + c x^{2})}^{10} = p_{0} + p_{1} x + p_{2} x^{2} + \dots$

${}^{10}C_{1} a^{9} \cdot b = p_{1} and {}^{10}C_{2} a^{8} \cdot b^{2} + {}^{10}C_{1} \cdot a^{9} \cdot c = p_{2}$

$\Rightarrow 10 b a^{9} = 20 and 45 b^{2} a^{8} + 10 c \cdot a^{9} = 210 \Rightarrow b a^{9} = 2$

$\Rightarrow$ $b = 2, a = 1$ $[∵ a, b \in ℕ]$

Now, $45 \times {(2)}^{2} + 10 \cdot c = 210$

$\Rightarrow 10 c = 210 - 180 \Rightarrow c = 3$

$∴$ $2 (a + b + c) = 2 (1 + 2 + 3) = 2 \times 6 = 12$

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