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If \(\left(1-x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}\), then \(2 a_2+3 a_3+4 a_4+\ldots+20 a_{20}=\)
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Verified Answer
The correct answer is:
20
\(\left(1-x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots \ldots+a_{20} x^{20}\) On differentiating
\(\begin{aligned}
& 10\left(1-x+x^2\right)^9 \cdot(-1+2 x)=a_1+2 a_2 x+3 a_3 x^2 \\
& +\ldots \ldots+20 a_{20} x^{19}
\end{aligned}\)
At \(x=1\)
\(\Rightarrow \quad a_1+2 a_2+3 a_3+\ldots \ldots+20 a_{20}=10\)
At \(x=0 \Rightarrow a_1=-10\)
So, \(2 a_2+3 a_3+\ldots \ldots+20 a_{20}=20\)
\(\begin{aligned}
& 10\left(1-x+x^2\right)^9 \cdot(-1+2 x)=a_1+2 a_2 x+3 a_3 x^2 \\
& +\ldots \ldots+20 a_{20} x^{19}
\end{aligned}\)
At \(x=1\)
\(\Rightarrow \quad a_1+2 a_2+3 a_3+\ldots \ldots+20 a_{20}=10\)
At \(x=0 \Rightarrow a_1=-10\)
So, \(2 a_2+3 a_3+\ldots \ldots+20 a_{20}=20\)
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