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STD 11 - 7. binomial theoram — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 7. binomial theoram4 Marks
MCQ
Let ${s_1} = \mathop \sum \limits_{j = 1}^{10} j\left( {j - 1} \right)\left( {\begin{array}{*{20}{c}}{10}\\j\end{array}} \right)\;,$$\;{s_2} = \mathop \sum \limits_{j = 1}^{10} j\;\left( {\begin{array}{*{20}{c}}{10}\\j\end{array}} \right)\;and,$${s_3} = \mathop \sum \limits_{j = 1}^{10} {j^2}\left( {\begin{array}{*{20}{c}}{10}\\j\end{array}} \right)\;,\;$

Statement $-1$:${s_3} = 55 \times {2^9}$

Statement $-2$: ${s_1} = 90 \times {2^8}\;$ and ${s_2} = 10 \times {2^8}$ 

  • A
    Statement $-1$ is true, Statement$-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$
  • B
    Statement $-1$ is true, Statement$-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$
  • C
    Statement $-1$ is false, Statement$-2$ is true
  • ✓
    Statement $-1$ is true, Statement$-2$ is false
✓

Answer

Correct option: D.
Statement $-1$ is true, Statement$-2$ is false
d
$S_{1} =\sum j(j-1)^{10} C_{j}$

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

$ = 9 \times 10\sum\limits_{j = 2}^{10} {^8{C_{j - 2}}}  = 90 \times {2^8}$

${S_2} = \sum\limits_{j = 1}^{10} j { \cdot ^{10}}{C_j} = 10\sum\limits_{j = 1}^{10} {^9{C_{j - 1}}}  = 10 \times {2^9}$

${S_3} = \sum\limits_{j = 1}^{10} {{j^2}} { \cdot ^{10}}{C_j} = \sum\limits_{j = 1}^{10} {(j(j - 1) + j)} { \cdot ^{10}}{C_j}$

$ = \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

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