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Binomial Coefficients

For r=0,1,....

Question

For $r = 0, 1, . . . ., 10$ , let $A_{r}, B_{r}$ and $C_{r}$ denote, respectively, the coefficient of $x^{r}$ in the expansions of ${(1 + x)}^{10}, {(1 + x)}^{20}$ and ${(1 + x)}^{30}$ . Then, $\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$ is equal to -

A

B_{10} - C_{10}

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B

A_{10} ({B_{10}}^{2} - C_{10} A_{10})

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C

0

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D

C_{10} - B_{10}

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Solution

The correct option is D $C_{10} - B_{10}$
Let $y = \sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$ .
Now, $\sum_{r = 1}^{10} A_{r} B_{r}$ = coefficient of $[{(1 + x)}^{10} {(1 + x)}^{20}]$ $- 1$
$\Rightarrow \sum_{r = 1}^{10} A_{r} B_{r} = C_{20} - 1 = C_{10} - 1$
And $\sum_{r = 1}^{10} {A_{r}}^{2} =$ coefficient of $x^{10}$ in $[{(1 + x)}^{10} {(1 + x)}^{10}]$ $- 1$
$\Rightarrow \sum_{r = 1}^{10} {A_{r}}^{2} = B_{10} - 1$
Therefore, $y = B_{10} (C_{10} - 1) - C_{10} (B_{10} - 1) = C_{10} - B_{10}$ .
Hence, option D is correct.

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Similar questions

r = 0, 1, 2,, . . . .10

A_{r}, B_{r}

C_{r}

denote respectively the coefficient of

x^{r}

in the expansions of

{(1 + x)}^{10}, {(1 + x)}^{20}

{(1 + x)}^{30}

\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})

r = 0, 1, . . ., 10

A_{r}, B_{r},

C_{r}

denote, respctively, the coefficient of

x^{r}

in the expansions of

(1 + x)^{10}, (1 + x)^{20}

(1 + x)^{30}

\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})

A_{r}, B_{r}, C_{r}

denotes the coefficients of

x^{r}

in the expansion of

(1 + x)^{10}, (x + 1)^{20}, (1 + x)^{30}

respectively, then the value of

10 \sum r = 1 A_{r} (B_{10} B_{r} - C_{10} A_{r})

r = 0, 1, . . ., 10

A_{r}, B_{r}

C_{r}

denote, respectively, the coefficient of

x^{r}

in the expansions of

(1 + x)^{10}, (1 + x)^{20}

(1 + x)^{30}

10 \sum r = 1 A_{r} (B_{10} B_{r} - C_{10} A_{r})

r = 0, 1, \dots

A_{r}, B_{r}

C_{r}

denote, respectively, the coefficient of

x^{r}

in the expansions of

(1 + x)^{10}, (1 + x)^{20}

(1 + x)^{30}

10 \sum r = 1 A_{r} (B_{10} B_{r} - C_{10} A_{r})

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