Question

Find the term independent of x in the expansion of (x + 1/x)10.​

Answer 1

Term independent of x index of the means index of the term should be zero.

first we will find the general term.

Tr+1= 10Cr(2x)^10-r(-1/x)^r

solve this and you will get index of x=10-2r

put 10-2r=0

r=5

if r=5then term will be 6th

put r=5 in the general term.

and you will get T6= -4032

Answer 2

❏Question:-

$\sf \red{ Find \: the \: term} \green{ \: independent \: of \: x }\\ \sf \purple{in \: the \: expansion \: of }\: (x+ \frac{1}{x} )^{10}</p><p> \:$

❏Answer:-

$\red{ \sf Term \: independent \: of \: x \: is \: } \sf 252 \: \: or \: ^{10}C_{5} \: \: \:$

❏To Find :-

→ Term independent of x .

❏ Explanation :-

We have expansion -

$\leadsto \sf { \bigg(x + \frac{1}{x} \bigg) }^{10} \\ \\ \sf \green{term \: independent \: of \: x }\: means \: the \: term \: \\ \sf \orange{which \: is \: not \: depend \: on \: }x \: ( {x}^{0}) \\ \bf \: hence \\ \\ \because \sf \red{ for \: coefficient \: of} \: x \: \: \blue{ \{ ^{n}C_{r} \: {a}^{n - r} {x}^{r} \} }\: \\ \\ \to \sf ^{10}C_{r} \: \: {x}^{(10 - r)} \: { \bigg( \frac{1}{x} \bigg)}^{r} \\ \\ \to \sf \: ^{10}C_{r} \: {x}^{(10 - r)} {x}^{( - r)} \\ \\ \to \sf \: ^{11}C_{r} \: \: {x}^{(10 - r - r)} \\ \\ \bf \red{ for \: term \:} \green{ independent \: of \: x} \\ \\ \to \sf 10 - r - r = 0 \\ \\ \to \sf \: 2r = 10 \\ \\ \to \boxed{ \sf r = 5} \\ \\ \sf hence \\ \\ \leadsto \sf ^{10}C_{5} \: {x}^{(10 - 5 - 5)} \\ \\ \to \sf ^{10}C_{5} \: {x}^{1} \\ \sf or \: \\ \to \sf \frac{10 \times 9 \times 8 \times 7 \times 6 }{5 \times 4 \times 3 \times 2 \times 1} \\ \\ \leadsto \sf \: 252 \\ \\ \sf Term \: independent \: of \: x \: is \: 252 \: \: or \: ^{10}C_{5} \: \:$

Hope it helps you .