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  2. Solve this problem - Binomial theorem and its simple applications - JEE Main-2
# Engineering

If ^{20}C_{1}+(2^{2}) ^{20}C_{2}+(3^{2}) + ^{20}C_{3} + \cdots \cdots

+(20^{2}) ^{20}C_{20}=A(2^{\beta }), then the ordered pair (A, \beta ) is equal to :

 

  • Option 1)

    (420, 19)

  • Option 2)

    (420, 18)

     

  • Option 3)

    (380, 18)

     

  • Option 4)

    (380, 19)

     

Answers (1)
V Vakul

^{20}C_{1}+(2^{2}) ^{20}C_{2}+(3^{2}) + ^{20}C_{3} + \cdots \cdots+(20^{2}) ^{20}C_{20}=A(2^{\beta })

LHS : 

\sum_{r=1}^{20}r^{2}\cdot (_{20}^{r}C\textrm{})

=>\sum_{r=1}^{20}r\cdot (r\cdot _{20}^{r}C\textrm{})

=>20\sum_{r=1}^{20}(r\cdot _{r-1}^{19}C\textrm{})

=>20\sum_{r=1}^{20}(r-1+1) _{r-1}^{19}C\textrm{}

=>20\sum_{r=1}^{20}(r-1) _{r-1}^{19}C\textrm{}+20\sum_{r=1}^{20}_{r-1}^{19}C\textrm{}

=>20\times 19\sum_{r=2}^{20}_{r-2}^{18}C\textrm{}+20\times2^{19}

=>20\times 19\times2^{18}+20\times2^{19}

=>20\times 2^{18}(19+2)

=>20\times21\times 2^{18}

=>420\times 2^{18}

Comparing LHS with RHS

A=420;\beta =18

 


Option 1)

(420, 19)

Option 2)

(420, 18)

 

Option 3)

(380, 18)

 

Option 4)

(380, 19)

 

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