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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)

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

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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Vakul

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