If ${ }^{20} \mathrm{C}_{\mathrm{r}}$ is the co-efficient of $\mathrm{x}^{\mathrm{r}}$ in the expansion of $(1+\mathrm{x})^{20}$, then the value of $\sum_{\mathrm{r}=0}^{20} \mathrm{r}^{2}{ }^{20} \mathrm{C}_{\mathrm{r}}$ is equal to

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Binomial Theorem

^{20} C_{r}

is the co-efficient of

x^{r}

in the expansion of

(1 + x)^{20}

, then the value of

\sum_{r = 0}^{20} r^{2}^{20} C_{r}

is equal to

Answer: (B)

Write

r^{2} = r (r - 1) + r

and use

r^{20} C_{r} = 20^{19} C_{r - 1}

together with

r (r - 1)^{20} C_{r} = 20 \cdot 19^{18} C_{r - 2}

. Then

r = 0 \sum 20 r^{20} C_{r} = 20 r = 1 \sum 20^{19} C_{r - 1} = 20 \cdot 2^{19},

r = 0 \sum 20 r (r - 1)^{20} C_{r} = 380 r = 2 \sum 20^{18} C_{r - 2} = 380 \cdot 2^{18} .

Adding the two,

r = 0 \sum 20 r^{2}^{20} C_{r} = 380 \cdot 2^{18} + 20 \cdot 2^{19} = 380 \cdot 2^{18} + 40 \cdot 2^{18} = 420 \times 2^{18} .

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option