The value of $-{ }^{15} \mathrm{C}_{1}+2 .{ }^{15} \mathrm{C}_{2}-3 .{ }^{15} \mathrm{C}_{3}+\ldots . .-15 .{ }^{15} \mathrm{C}_{15}+{ }^{14} \mathrm{C}_{1}+{ }^{14} \mathrm{C}_{3}+{ }^{14} \mathrm{C}_{5}+\ldots .+{ }^{14} \mathrm{C}_{11}$ is :

Binomial Theorem — JEE Main · 2021 · 24 Feb 2021 · Shift 1 Question with Solution

The value of

-^{15} C_{1} + 2.^{15} C_{2} - 3.^{15} C_{3} + \dots .. - 15.^{15} C_{15} +^{14} C_{1} +^{14} C_{3} +^{14} C_{5} + \dots . +^{14} C_{11}

is :

Answer: (A)

First part:

\sum_{r = 1}^{15} (- 1)^{r} r (r 15)

.

Using the identity

r (r n) = n (r - 1 n - 1)

r = 1 \sum 15 (- 1)^{r} r (r 15) = 15 r = 1 \sum 15 (- 1)^{r} (r - 1 14) = 15 k = 0 \sum 14 (- 1)^{k + 1} (k 14) = - 15 (1 - 1)^{14} = 0

Wait, that's not quite right. Let me reconsider. Actually,

\sum_{r = 0}^{n} (- 1)^{r} (r n) = 0

for

n > 0

. So:

r = 1 \sum 15 (- 1)^{r} r (r 15) = 15 k = 0 \sum 14 (- 1)^{k + 1} (k 14) = - 15 \cdot 0 = 0

Second part:

\sum_{k = 0}^{5} (2 k + 1 14)

(odd binomial coefficients).

Using

(1 + 1)^{14} = 2^{14}

and

(1 - 1)^{14} = 0

, the sum of odd coefficients equals the sum of even coefficients, each being

2^{13}

.

But we need

(1 14) + (3 14) + \dots + (11 14)

(6 terms out of 7 odd positions).

Total odd sum =

2^{13}

, and

(13 14) = 14

is the missing term.

So the sum is

2^{13} - 14

.

Total:

0 + (2^{13} - 14) = 2^{13} - 14

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option