Question

Previous year Question of jee mains

mathamatics shift :1
24 feb 2021 ​

Answer 1

Answer:

The value of first expansion is 0

and value of second exapnsion is 2¹³-14

hence,

the answer is 0+2¹³-14

= 2¹³-14

Answer 2

$\large\underline{\sf{Solution-}}$

We know,

$\rm :\longmapsto\:In \: Binomial \: expansion \: of \: {(1 + x)}^{n}, \: we \: have$

$\boxed{\rm{^nC_ 0 + ^nC_ 1 + ^nC_ 2 + - - - + ^nC_ n = {2}^{n}}}$

$\boxed{\rm{^nC_ 0 - ^nC_ 1 + ^nC_ 2 + - - - {( - 1)}^{n} \: ^nC_ n = 0}}$

$\boxed{ \rm{ \:^nC_ 0 = ^nC_ 2 = ^nC_ 4 - - - = ^nC_ 1 + ^nC_ 3 + ^nC_ 5 + - - - = {2}^{n - 1}}}$

Also, we know that

$\boxed{ \sf{ \:^nC_ r \: = \: \frac{n}{r} \: \: ^{n - 1}C_{r - 1}}}$

Now,

Let's solve the problem now!!

Consider,

$\rm :\longmapsto\:^{14}C_1 + ^{14}C_3 + ^{14}C_5 + - - - + ^{14}C_{11}$

$\rm \: = \:^{14}C_1 + ^{14}C_3 + ^{14}C_5 + - - + ^{14}C_{11} + ^{14}C_{13} - ^{14}C_{13}$

$\rm \: =( \:^{14}C_1 + ^{14}C_3 + ^{14}C_5 + - - + ^{14}C_{11} + ^{14}C_{13}) - ^{14}C_{13}$

$\rm \:  =  \:  {2}^{14 - 1} - ^{14}C_1 \: \: \: \: \: \purple{\{ \because \: ^{14}C_{13} = ^{14}C_1 \}}$

$\rm \:  =  \: {2}^{13} - 14$

Hence,

$\bf :\longmapsto\:^{14}C_1 + ^{14}C_3 + ^{14}C_5 + -- + ^{14}C_{11} = {2}^{13} - 14$

Now, Consider

$\rm :\longmapsto\: - ^{15}C_1 + 2.^{15}C_2 - 3.^{15}C_3 + - - - 15.^{15}C_{15}$

Now,

Above terms can be rewritten as

$\rm :\longmapsto\:^{15}C_1 = \dfrac{15}{1} \: ^{14}C_0$

$\rm :\longmapsto\:^{15}C_2 = \dfrac{15}{2} \: ^{14}C_1$

$\rm :\longmapsto\:^{15}C_3 = \dfrac{15}{3} \: ^{14}C_2$

.

.

.

$\rm :\longmapsto\:^{15}C_{15} = \dfrac{15}{15} \: ^{14}C_{14}$

So,

$\rm :\longmapsto\: - ^{15}C_1 + 2.^{15}C_2 - 3.^{15}C_3 + - - - 15.^{15}C_{15}$

$\rm = - 15.^{14}C_0 + 2. \dfrac{15}{2} \: ^{14}C_1 - 3.\dfrac{15}{3}^{14}C_2 + - - - 15.^{14}C_{14}$

$\rm \:  =  \:  -15 \: ^{14}C_0 + 15.^{14}C_1 - 15.^{14}C_2 - - 15.^{14}C_{14}$

$\rm \:  =  \:  -15( \: ^{14}C_0 -^{14}C_1 + ^{14}C_2 - - - + ^{14}C_{14})$

$\rm \:  =  \:  \: - 15 \times 0$

$\rm \:  =  \:  \: 0$

Hence,

$\rm :\longmapsto\: - ^{15}C_1 + 2.^{15}C_2 - 3.^{15}C_3 + - - - 15.^{15}C_{15} = 0$

So,

$\rm :\longmapsto\:- ^{15}C_1 + 2.^{15}C_2 - 3.^{15}C_3 + - - - 15.^{15}C_{15} + ^{14}C_1 + ^{14}C_3 + ^{14}C_5 + - - - + ^{14}C_{11}$

$\bf \:  =  \: {2}^{13} - 14$

Hence, Option (2) is correct.