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in Binomial Theorem by (50.8k points)
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Find the middle terms in the expansion of:

(i) (3x – x3/6)9

(ii) (2x2 – 1/x)7

(iii) (3x – 2/x2)15

(iv) (x4 – 1/x3)11

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Best answer

(i) (3x – x3/6)9

Now, we have,

(3x – x3/6)9 where, n = 9 (odd number)

Therefore the middle terms are ((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 and 6th.

Then,

T5 = T4+1

Thus, the middle term are 189/8 x17 and -21/16 x19.

(ii) (2x2 – 1/x)7

Now we have,

(2x2 – 1/x)7 where, n = 7 (odd number)

Therefore the middle terms are ((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 and 5th.

Then,

Thus, the middle term are -560x5 and 280x2.

(iii) (3x – 2/x2)15

We have,

(3x – 2/x2)15 where, n = 15 (odd number)

Therefore the middle terms are ((n + 1)/2) = ((15 + 1)/2) = 16/2 = 8 and

((n + 1)/2 + 1) = ((15 + 1)/2 + 1) = (16/2 + 1) = (8 + 1) = 9

The terms are 8th and 9th.

Now,

Thus, the middle term are (-6435×38×27)/x6 and (6435×37×28)/x9.

(iv) (x4 – 1/x3)11

Here, we have,

(x4 – 1/x3)11

where, n = 11 (odd number)

Therefore the middle terms are ((n + 1)/2) = ((11 + 1)/2) = 12/2 = 6 and

((n + 1)/2 + 1) = ((11 + 1)/2 + 1) = (12/2 + 1) = (6 + 1) = 7

The terms are 6th and 7th.

Now,

Thus, the middle term are -462x9 and 462x2.

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