माना S_{1}=sum_{j=1}^{10} j(j-1)^{10} C_{j}, | vedclass

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Question

$S_{1} =\sum j(j-1)^{10} C_{j}$

$=\sum j(j-1) \cdot \frac{10(10-1)}{(j-1)} \cdot^{8} C_{j-2}$

$ = 9 	imes 10\sum\limits_{j = 2}^{10} {^8{C_{j -

Vedclass · Accepted Answer

कथन $1$ सत्य है, कथन $2$ असत्य है।

Vedclass · Answer

$S_{1} =\sum j(j-1)^{10} C_{j}$

$=\sum j(j-1) \cdot \frac{10(10-1)}{(j-1)} \cdot^{8} C_{j-2}$

$ = 9 	imes 10\sum\limits_{j = 2}^{10} {^8{C_{j - 2}}}  = 90 	imes {2^8}$

${S_2} = \sum\limits_{j = 1}^{10} j { \cdot ^{10}}{C_j} = 10\sum\limits_{j = 1}^{10} {^9{C_{j - 1}}}  = 10 	imes {2^9}$

${S_3} = \sum\limits_{j = 1}^{10} {{j^2}} { \cdot ^{10}}{C_j} = \sum\limits_{j = 1}^{10} {(j(j - 1) + j)} { \cdot ^{10}}{C_j}$

$ = \sum\limits_{j = 1}^{10} {j{{(j - 1)}^{10}}{C_j}}  + \sum\limits_{j = 1}^{10} {j{.^{10}}{C_j}} $

$=90 \cdot 2^{8}+10 \cdot 2^{9}=(45+10) 2^{9}=55 \cdot 2^{9}$

Then statement $-1$ is true and statement $-2$ is false

माना $S_{1}=\sum_{j=1}^{10} j(j-1)^{10} C_{j}, S_{2}=\sum_{j=1}^{10} j^{10} C_{j}$

और $S_{3}=\sum_{j=1}^{10} j^{210} C_{j}$

कथन $1: S_{3}=55 \times 2^{9}$

कथन $2: S_{1}=90 \times 2^{8}$ और $S_{2}=10 \times 2^{8}$

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