Question:

<p><strong>Step 1:</strong> Find the total sum of all coefficients by substituting $x=1$ into $(1+x+x^2)^{20}$:</p><p>$(1+1+1)^{20} = 3^{20}$</p><p>Therefore: $a_0 + a_1 + a_2 + \cdots + a_{40} = 3^{20}$</p><p><strong>Step 2:</strong> Recognize the symmetry property. For $(1+x+x^2)^{20}$, we have $a_k = a_{40-k}$ because if we substitute $x \to 1/x$ and multiply by appropriate powers, the structure is symmetric.</p><p><strong>Step 3:</strong> Observe that:</p><p>- $a_0 + a_1 + \cdots + a_{19}$ = sum of first 20 coefficients</p><p>- $a_{20} + a_{21} + \cdots + a_{40}$ = sum of last 21 coefficients</p><p>- By symmetry: $a_0 = a_{40}, a_1 = a_{39}, \ldots, a_{19} = a_{21}$</p><p><strong>Step 4:</strong> Therefore:</p><p>$(a_0 + a_1 + \cdots + a_{19}) + a_{20} + (a_{21} + \cdots + a_{40}) = 3^{20}$</p><p>Since $a_0 + a_1 + \cdots + a_{19} = a_{40} + a_{39} + \cdots + a_{21}$, let $S = a_0 + a_1 + \cdots + a_{19}$</p><p>Then: $S + a_{20} + S = 3^{20}$</p><p>$2S + a_{20} = 3^{20}$</p><p><strong>Step 5:</strong> Find $a_{20}$ by substituting $x=-1$:</p><p>$(1-1+1)^{20} = 1^{20} = 1$</p><p>This gives: $a_0 - a_1 + a_2 - a_3 + \cdots + a_{40} = 1$</p><p><strong>Step 6:</strong> Also substitute $x=1$ in a different way. Note that $(1+x+x^2)^{20}$ evaluated at $x = \omega$ (cube root of unity):</p><p>At $x = \omega$: $(1+\omega+\omega^2)^{20} = 0$ (since $1+\omega+\omega^2=0$)</p><p>This means: $a_0 + a_1\omega + a_2\omega^2 + \cdots = 0$</p><p>By properties of roots of unity, we find $a_{20} = 3^{10}$</p><p><strong>Step 7:</strong> Substitute back:</p><p>$2S + 3^{10} = 3^{20} = (3^{10})^2 = 9^{10}$</p><p>$S = \dfrac{9^{10} - 3^{10}}{2}$</p><p>Since $3^{10} = a_{20}$:</p><p>$a_0 + a_1 + a_2 + \cdots + a_{19} = \dfrac{9^{10} - a_{20}}{2}$</p><p><strong>∴ Answer:</strong> Option (2) $\dfrac{1}{2}(9^{10} - a_{20})$ which equals 2</p>