# Find the coefficient of 1x17 in the expansion of (x4-1x3)15 - Mathematics

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Sum

Find the coefficient of 1/x^17 in the expansion of (x^4 - 1/x^3)^15

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#### Solution

The given expression is (x^4 - 1/x^3)^15

General Term "T"_(r + 1) = ""^n"C"_r x^(n - r) y^r

= ""^15"C"_r  (x^4)^(15 - r) (- 1/x^3)^r

= ""^15"C"_r  (x)^(60 - 4r) (-1)^r * 1/x^(3r)

= ""^15"C"_r (-1)^r * 1/(x^(3r - 60 + 4r))

= ""^15"C"_r (-1)^r * 1/(x^(7r - 60))

To find the coefficient of 1/x^17

Put 7r – 60 = 17

⇒ 7r = 60 + 17

⇒ 7r = 77

∴ r = 11

Putting the value of r in the above expression, we get

= ""^15"C"_11 (-1)^11 * 1/x^17

= - ""^15"C"_4 * 1/x^17

= - (15 xx 14 xx 13 xx 12)/(4 xx 3 xx 2 xx 1) * 1/x^17

= - 1365 * 1/x^17

Hence, the coefficient of 1/x^17 = – 1365

Concept: General and Middle Terms
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Chapter 8: Binomial Theorem - Exercise [Page 143]

#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Exercise | Q 7 | Page 143

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