Question -
Answer -
(i) (x – 1/x)10
We have,
(x – 1/x)10 where,n = 10 (even number)
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle termis -252.
(ii) (1 – 2x + x2)n
We have,
(1 – 2x + x2)n =(1 – x)2n where, n is an even number.
So the middle term is(2n/2 + 1) = (n + 1)th term.
Now,
Tn = Tn+1
= 2nCn (-1)n (x)n
= (2n)!/(n!)2 (-1)n xn
Hence, the middle termis (2n)!/(n!)2 (-1)n xn.
(iii) (1 + 3x + 3x2 +x3)2n
We have,
(1 + 3x + 3x2 +x3)2n = (1 + x)6n where, n is aneven number.
So the middle term is(n/2 + 1) = (6n/2 + 1) = (3n + 1)th term.
Now,
T2n =T3n+1
= 6nC3n x3n
= (6n)!/(3n!)2 x3n
Hence, the middle termis (6n)!/(3n!)2 x3n.
(iv) (2x – x2/4)9
We have,
(2x – x2/4)9 where,n = 9 (odd number)
So the middle termsare ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and
((n+1)/2 + 1) =((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and6th.
Now,
T5 = T4+1
And,
T6 = T5+1
Hence, the middle termis 63/4 x13 and -63/32 x14.
(v) (x – 1/x)2n+1
We have,
(x – 1/x)2n+1 where,n = (2n + 1) is an (odd number)
So the middle termsare ((n+1)/2) = ((2n+1+1)/2) = (2n+2)/2 = (n + 1) and
((n+1)/2 + 1) =((2n+1+1)/2 + 1) = ((2n+2)/2 + 1) = (n + 1 + 1) = (n + 2)
The terms are (n + 1)th and(n + 2)th.
Now,
Tn = Tn+1
And,
Tn+2 =Tn+1+1
Hence, the middle termis (-1)n.2n+1Cn x and (-1)n+1.2n+1Cn (1/x).
(vi) (x/3 + 9y)10
We have,
(x/3 + 9y)10 where,n = 10 is an even number.
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. i.e., 6th term.
Now,
T6 = T5+1
Hence, the middle termis 61236x5y5.
(vii) (3 – x3/6)7
We have,
(3 – x3/6)7 where,n = 7 (odd number).
So the middle termsare ((n+1)/2) = ((7+1)/2) = 8/2 = 4 and
((n+1)/2 + 1) =((7+1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5
The terms are 4th and5th.
Now,
T4 = T3+1
= 7C3 (3)7-3 (-x3/6)3
= -105/8 x9
And,
T5 = T4+1
= 9C4 (3)9-4 (-x3/6)4
Hence, the middleterms are -105/8 x9 and 35/48 x12.
(viii) (2ax – b/x2)12
We have,
(2ax – b/x2)12 where,n = 12 is an even number.
So the middle term is(n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. i.e., 7th term.
Now,
T7 = T6+1
Hence, the middle termis (59136a6b6)/x6.
(ix) (p/x + x/p)9
We have,
(p/x + x/p)9 where,n = 9 (odd number).
So the middle termsare ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and
((n+1)/2 + 1) =((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and6th.
Now,
T5 = T4+1
And,
T6 = T5+1
= 9C5 (p/x)9-5 (x/p)5
Hence, the middleterms are 126p/x and 126x/p.
(x) (x/a – a/x)10
We have,
(x/a – a/x) 10 where,n = 10 (even number)
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle termis -252.