Let (1+2 x)^{20}=a_0+a₁ x+a₂ x^2+ldots+a_{20} x^{20} .Then 3 a

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7.Binomial Theorem

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

$\frac{5 \cdot 3^{20}-3}{2}$

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

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

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

(KVPY-2009)

Solution

(c)

We have,

$(1+2 x)^{20}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$

Put $x=1$,

$3^{20}=a_0+a_1+a_2+\ldots+a_{20}$

Put $x=-1$,

$1=a_0-a_1+a_2-a_3+\ldots+a_{20} \ldots \text { (ii) }$

On adding Eqs.$(i)$ and $(ii)$, we get $\frac{3^{20}+1}{2}=a_0+a_2+a_4+\ldots+a_{20}$

On subtracting Eq.$(ii)$ from Eq.$(i)$,

we get $\frac{3^{20}-1}{2}=a_1+a_3+a_5+\ldots+a_{19}$

Now, we have

$3 a_0+2 a_1+3 a_2+2 a_3+\ldots+2 a_{19}+3 a_{20}$

$=3\left(a_0+a_2+a_4+\ldots+a_{20}\right)$

$+3\left(\frac{3^{20}+2\left(a_1+a_3+\ldots+a_{19}\right)}{2}\right)+2\left(\frac{3^{20}-1}{2}\right)$

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

Standard 11

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Solution

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