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

normal

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

Mathematics

Co-efficient of $\alpha ^t$ in the expansion of,

$(\alpha + p)^{m – 1} + (\alpha + p)^{m – 2} (\alpha + q) + (\alpha + p)^{m – 3} (\alpha + q)^2 + …… (\alpha + q)^{m – 1}$

where $\alpha \ne – q$ and $p \ne q$ is :

normal

If the sum of the coefficients in the expansion of ${(x – 2y + 3z)^n}$ is $128$ then the greatest coefficient in the expansion of ${(1 + x)^n}$ is

medium

If $r,k,p \in W,$ then $\sum\limits_{r + k + p = 10} {{}^{30}{C_r} \cdot {}^{20}{C_k} \cdot {}^{10}{C_p}} $ is equal to –

normal

A possible value of $^{\prime}x^{\prime}$, for which the ninth term in the expansion of $\left\{3^{\log _{3} \sqrt{25^{x-1}+7}}+3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)}\right\}^{10}$ in the increasing powers of $3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)}$ is equal to $180$ , is:

hard

(JEE MAIN-2021)

The sum of coefficients of integral power of $x$ in the binomial expansion ${\left( {1 – 2\sqrt x } \right)^{50}}$ is :

hard

(JEE MAIN-2015)

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

Solution

Similar Questions

Co-efficient of $\alpha ^t$ in the expansion of, $(\alpha + p)^{m – 1} + (\alpha + p)^{m – 2} (\alpha + q) + (\alpha + p)^{m – 3} (\alpha + q)^2 + …… (\alpha + q)^{m – 1}$ where $\alpha \ne – q$ and $p \ne q$ is :

If the sum of the coefficients in the expansion of ${(x – 2y + 3z)^n}$ is $128$ then the greatest coefficient in the expansion of ${(1 + x)^n}$ is

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Co-efficient of $\alpha ^t$ in the expansion of,

$(\alpha + p)^{m – 1} + (\alpha + p)^{m – 2} (\alpha + q) + (\alpha + p)^{m – 3} (\alpha + q)^2 + …… (\alpha + q)^{m – 1}$

where $\alpha \ne – q$ and $p \ne q$ is :