Let (1+2x)^{20} = a_0 + a_1x + a_2x^2 + dots | vedclass

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(C) Given $(1+2x)^{20} = a_0 + a_1x + a_2x^2 + \dots + a_{20}x^{20}$.Putting $x=1$,we get $3^{20} = a_0 + a_1 + a_2 + \dots + a_{20} \dots (i)$.Puttin

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$\frac{5 \cdot 3^{20}+1}{2}$

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(C) Given $(1+2x)^{20} = a_0 + a_1x + a_2x^2 + \dots + a_{20}x^{20}$.Putting $x=1$,we get $3^{20} = a_0 + a_1 + a_2 + \dots + a_{20} \dots (i)$.Putting $x=-1$,we get $(-1)^{20} = a_0 - a_1 + a_2 - a_3 + \dots + a_{20}$,so $1 = a_0 - a_1 + a_2 - a_3 + \dots + a_{20} \dots (ii)$.Adding $(i)$ and $(ii)$,we get $a_0 + a_2 + a_4 + \dots + a_{20} = \frac{3^{20}+1}{2}$.Subtracting $(ii)$ from $(i)$,we get $a_1 + a_3 + a_5 + \dots + a_{19} = \frac{3^{20}-1}{2}$.We need to evaluate $S = 3a_0 + 2a_1 + 3a_2 + 2a_3 + \dots + 2a_{19} + 3a_{20}$.This can be written as $S = 3(a_0 + a_2 + \dots + a_{20}) + 2(a_1 + a_3 + \dots + a_{19})$.Substituting the values,$S = 3 \left( \frac{3^{20}+1}{2} \right) + 2 \left( \frac{3^{20}-1}{2} \right)$.$S = \frac{3 \cdot 3^{20} + 3 + 2 \cdot 3^{20} - 2}{2} = \frac{5 \cdot 3^{20} + 1}{2}$.

Let $(1+2x)^{20} = a_0 + a_1x + a_2x^2 + \dots + a_{20}x^{20}$. Then $3a_0 + 2a_1 + 3a_2 + 2a_3 + 3a_4 + 2a_5 + \dots + 2a_{19} + 3a_{20}$ equals

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