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Mathematical Induction and Binomial Theorem question

2010 · Shift 2 · Q25

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JEE Advanced
Mathematics
Mathematical Induction and Binomial Theorem
2010 · Shift 2 · Q25

Mathematical Induction and Binomial Theorem question

2010 · Shift 2 · Q25

JEE AdvancedMathematicsMathematical Induction and Binomial TheoremMCQ+4 / −1

For

r = 0,\,1,....,

let

{A_r},\,{B_r}

and

{C_r}

denote, respectively, the coefficient of

{X^r}

in the expansions of

{\left( {1 + x} \right)^{10}},{\left( {1 + x} \right)^{20}}

and

{\left( {1 + x} \right)^{30}}.

Then

\sum\limits_{r = 1}^{10} {{A_r}\left( {{B_{10}}{B_r} - {C_{10}}{A_r}} \right)}

is equal to

A
$\left( {{B_{10}} - {C_{10}}} \right)$
B
${A_{10}}\left( {{B^2}_{10}{C_{10}}{A_{10}}} \right)$
C
$0$
D
${{C_{10}} - {B_{10}}}$

View written solutionFree

Correct answer: D

Interpret the coefficients

Given:

$A_r$ is the coefficient of $x^r$ in $(1+x)^{10}$ , so $A_r = \binom{10}{r}.$
$B_r$ is the coefficient of $x^r$ in $(1+x)^{20}$ , so $B_r = \binom{20}{r}.$
$C_r$ is the coefficient of $x^r$ in $(1+x)^{30}$ , so $C_r = \binom{30}{r}.$

We need to evaluate $S=\sum_{r=1}^{10} A_r\left(B_{10}B_r-C_{10}A_r\right).$

Substitute the meanings of $A_r,B_r,C_r$ : $S=B_{10}\sum_{r=1}^{10}A_rB_r-C_{10}\sum_{r=1}^{10}A_r^2.$

So we must compute the two sums: $\sum_{r=1}^{10}A_rB_r \quad \text{and} \quad \sum_{r=1}^{10}A_r^2.$

Compute $\sum_{r=1}^{10} A_r B_r$

Using $A_r=\binom{10}{r}, \qquad B_r=\binom{20}{r},$ we get $\sum_{r=0}^{10}A_rB_r=\sum_{r=0}^{10}\binom{10}{r}\binom{20}{r}.$

Now use the identity $\sum_{r=0}^{n}\binom{n}{r}\binom{m}{r}=\binom{n+m}{n}.$

Hence, $\sum_{r=0}^{10}\binom{10}{r}\binom{20}{r}=\binom{30}{10}=C_{10}.$

Therefore, $\sum_{r=1}^{10}A_rB_r=C_{10}-A_0B_0=C_{10}-1,$ since $A_0=B_0=1$ .

Compute $\sum_{r=1}^{10} A_r^2$

Using $A_r=\binom{10}{r}$ , $\sum_{r=0}^{10}A_r^2=\sum_{r=0}^{10}\binom{10}{r}^2.$

Use the standard identity $\sum_{r=0}^{n}\binom{n}{r}^2=\binom{2n}{n}.$

So, $\sum_{r=0}^{10}\binom{10}{r}^2=\binom{20}{10}=B_{10}.$

Hence, $\sum_{r=1}^{10}A_r^2=B_{10}-A_0^2=B_{10}-1,$ since $A_0=1$ .

Substitute back into $S$

Now, $S=B_{10}(C_{10}-1)-C_{10}(B_{10}-1).$

Expand: $S=B_{10}C_{10}-B_{10}-C_{10}B_{10}+C_{10}.$

The middle terms cancel: $S=C_{10}-B_{10}.$

Match with options

Thus the required value is $\boxed{C_{10}-B_{10}}.$

This is Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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