For $\mathrm{r}=0,1, \ldots, 10$, let $\mathrm{A}_{\mathrm{r}}, \mathrm{B}_{\mathrm{r}}$ and $\mathrm{C}_{\mathrm{r}}$ denote, respectively, the coefficient of $\mathrm{x}^{\mathrm{r}}$ in the expansions of $(1+\mathrm{x})^{10},(1+\mathrm{x})^{20}$ and $(1+\mathrm{x})^{30}$. Then \[ \sum_{r=1}^{10} A_{r}\left(B_{10} B_{r}-C_{10} A_{r}\right) \] is equal to

(D) $\mathrm{C}_{10}-\mathrm{B}_{10}$

(A) $\mathrm{B}_{10}-\mathrm{C}_{10}$

(B) $\mathrm{A}_{10}\left(\mathrm{~B}_{10}^{2}-\mathrm{C}_{10} \mathrm{~A}_{10}\right)$

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Binomial Theorem

For

r = 0, 1, \dots, 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

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

is equal to

Answer: (D)

We have

A_{r} = (r 10)

B_{r} = (r 20)

C_{r} = (r 30)

.

The sum can be written as:

r = 1 \sum 10 A_{r} B_{10} B_{r} - r = 1 \sum 10 C_{10} A_{r}^{2} = B_{10} r = 1 \sum 10 A_{r} B_{r} - C_{10} r = 1 \sum 10 A_{r}^{2}

Using Vandermonde's identity,

\sum_{r = 0}^{10} A_{r} B_{r}

equals the coefficient of

x^{10}

(1 + x)^{30}

, which is

C_{10}

. Thus

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

.

Similarly,

\sum_{r = 0}^{10} A_{r}^{2}

equals the coefficient of

x^{10}

(1 + x)^{20}

, which is

B_{10}

. Thus

\sum_{r = 1}^{10} A_{r}^{2} = B_{10} - 1

.

Therefore:

B_{10} (C_{10} - 1) - C_{10} (B_{10} - 1) = B_{10} C_{10} - B_{10} - C_{10} B_{10} + C_{10} = C_{10} - B_{10}

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Single correct option