2.{}^{20}{C_0} + 5.{}^{20}{C₁} + 8.{}^{20}{C₂} + 11.{}^{20}{C₃} + ......62.{}^{20}{C

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7.Binomial Theorem

hard

$2.{}^{20}{C_0} + 5.{}^{20}{C_1} + 8.{}^{20}{C_2} + 11.{}^{20}{C_3} + ......62.{}^{20}{C_{20}}$ is equal to

${2^{23}}$

${2^{26}}$

${2^{24}}$

${2^{25}}$

(JEE MAIN-2019)

Solution

$2 \cdot^{20} \mathrm{C}_{0}+5 \cdot^{20} \mathrm{C}_{1}+8 \cdot^{20} \mathrm{C}_{2}+11 \cdot^{20} \mathrm{C}_{3}+\ldots \ldots+62 \cdot^{20} \mathrm{C}_{20} 2$

$ = \sum\limits_{r = 0}^{20} {(3r + 2){\,^{20}}} {C_r}$

$ = 3\sum\limits_{r = 0}^{20} r { \cdot ^{20}}{C_r} + 2\sum\limits_{r = 0}^{20} {{\,^{20}}} {C_r}$

$ = 3\sum\limits_{r = 0}^{20} {r\left( {\frac{{20}}{r}} \right)} {\,^{19}}{C_{r – 1}} + {2.2^{20}}$

$=60.2^{19}+2.2^{20}=2^{25}$

Standard 11

Mathematics

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

normal

(KVPY-2009)

$\sum\limits_{n = 0}^4 {{{\left( {1009 – 2n} \right)}^4}\left( \begin{gathered}
4 \hfill \\
n \hfill \\
\end{gathered} \right)} {\left( { – 1} \right)^n}$ is

normal

If $n$ is an integer greater than $1$, then $a{ – ^n}{C_1}(a – 1){ + ^n}{C_2}(a – 2) + …. + {( – 1)^n}(a – n) = $

medium

(IIT-1972)

If $\frac{1}{n+1}{ }^n C_n+\frac{1}{n}{ }^n C_{n-1}+\ldots+\frac{1}{2}{ }^{ n } C _1+{ }^{ n } C _0=\frac{1023}{10}$ then $n$ is equal to

hard

(JEE MAIN-2023)

The number of terms in the expansion of $(1 +x)^{101} (1 +x^2 – x)^{100}$ in powers of $x$ is

hard

(JEE MAIN-2014)

$2.{}^{20}{C_0} + 5.{}^{20}{C_1} + 8.{}^{20}{C_2} + 11.{}^{20}{C_3} + ......62.{}^{20}{C_{20}}$ is equal to

Solution

Similar Questions

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

$\sum\limits_{n = 0}^4 {{{\left( {1009 – 2n} \right)}^4}\left( \begin{gathered} 4 \hfill \\ n \hfill \\ \end{gathered} \right)} {\left( { – 1} \right)^n}$ is

If $n$ is an integer greater than $1$, then $a{ – ^n}{C_1}(a – 1){ + ^n}{C_2}(a – 2) + …. + {( – 1)^n}(a – n) = $

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The number of terms in the expansion of $(1 +x)^{101} (1 +x^2 – x)^{100}$ in powers of $x$ is

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$\sum\limits_{n = 0}^4 {{{\left( {1009 – 2n} \right)}^4}\left( \begin{gathered}
4 \hfill \\
n \hfill \\
\end{gathered} \right)} {\left( { – 1} \right)^n}$ is