For r -0-1- - 10- let A r - B r- and C r denote- respctively- the coefficient of x r in the expansions of 1- x 10-1- x 20 and 1- x 30- Then - r -110 A r B 10 B r - C 10 A r is equal toA- B10-C10B- 0C- C 10- B 10D- A 10 B 102 C 10 A 10

The correct option is C C_10 B_10Clearly A_r = ^10C_r, B_r = ^20C_r, C_r = ^30C_r, Now ∑_r=1^10^10C_r(^20C_10 ^20C_r ^30C_10 ^10C_r ) = ^20C_10∑_r=1^10 ^10C ...

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Byju's Answer

Standard XI

Mathematics

Sum of product of binomial coefficients

For r =0,1, …...

Question

For $r = 0, 1, . . ., 10$ , let $A_{r}, B_{r},$ and $C_{r}$ denote, respctively, 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

B_{10} - C_{10}

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

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C_{10} - B_{10}

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Solution

The correct option is C $C_{10} - B_{10}$
$Clearly A_{r} =^{10} C_{r}, B_{r} =^{20} C_{r}, C_{r} =^{30} C_{r}, Now \sum_{r = 1}^{10}^{10} C_{r} (^{20} C_{10}^{20} C_{r} -^{30} C_{10}^{10} C_{r}) =^{20} C_{10} \sum_{r = 1}^{10}^{10} C_{r}^{20} C_{r} -^{30} C_{10} \sum_{r = 1}^{10}^{10} C_{r} \times^{10} C_{r} =^{20} C_{10} (^{10} C_{1}^{20} C_{1} +^{10} C_{2}^{20} C_{2} + . . . +^{10} C_{10}^{20} C_{10}) =^{30} C_{10} (^{10} C_{1} \times^{10} C_{1} +^{10} C_{2} \times^{10} C_{2} + . . . +^{10} C_{10}^{10} C_{10}) . . . (1)$
Now expanding $(1 + x)^{10}$ and $(1 + x)^{20}$ by binomial theorem and comparing the coefficients of $x^{20}$ in their product, on both sides, we get
$^{10} C_{0}^{20} C_{0} +^{10} C_{1}^{20} C_{1} +^{10} C_{2}^{20} C_{2} + . . . +^{10} C_{10}^{20} C_{10}$
= coeff of $x^{20}$ in $(1 + x)^{30} =^{30} C_{20} =^{30} C_{10} ∴^{10} C_{1}^{20} C_{1} +^{10} C_{2}^{20} C_{2} + . . . +^{10} C_{10}^{20} C_{10} =^{30} C_{10} - 1$
Again expending $(1 + x^{10})$ and $(x + 1)^{10}$ by binomial theorem and comparing the coefficients of $x^{10}$ in their product on both sides, we get
$∴ (^{10} C_{0})^{2} (^{10} C_{1})^{2} + (^{10} C_{2})^{2} + . . . + (^{10} C_{10})^{2}$
= coeff of $x^{10}$ in $(1 + x)^{20} =^{20} C_{10} ∴ (^{10} C_{1})^{2} + (^{10} C_{2})^{2} + . . . + (^{10} C_{10})^{2} =^{20} C_{10} - 1$
Substituting these values in equation (1), we get
$^{20} C_{10} (^{30} C_{10} - 1) -^{30} C_{10} (^{20} C_{10} - 1) =^{30} C_{10} -^{20} C_{10} = C_{10} - B_{10}$

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Similar questions

Q. For

r = 0, 1, . . ., 10

, let

A_{r}, B_{r}

and

C_{r}

denote, respectively, the coefficient of

x^{r}

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

and

(1 + x)^{30}

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

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

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For r=0,1,...,10, let Ar, Br, and Cr denote, respctively, the coefficient of xr in the expansions of (1+x)10, (1+x)20 and (1+x)30. Then ∑10r=1Ar(B10Br−C10Ar)is equal to

For $r = 0, 1, . . ., 10$ , let $A_{r}, B_{r},$ and $C_{r}$ denote, respctively, 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