wiz-icon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
Question

Find the coefficient of:
(i) x10 in the expansion of 2x2-1x20

(ii) x7 in the expansion of x-1x240

(iii) x-15 in the expansion of 3x2-a3x310

(iv) x9 in the expansion of x2-13x9

(v) xm in the expansion of x+1xn

(vi) x in the expansion of 1-2x3+3x5 1+1x8.

(vii) a5b7 in the expansion of a-2b12.

(viii) x in the expansion of 1-3x+7x2 1-x16.

Open in App
Solution

(i) Suppose x10 occurs in the (r + 1)th term in the given expression.

Then, we have:

Tr+1=Crn xn-r ar

Here,
Tr+1=Cr20(2x2)20-r -1xr =(-1)r Cr20220-r x40-2r-rFor this term to contain x10, we must have:40-3r =103r=30r=10 Coefficient of x10 = (-1)10 C1020220-10 =C1020210

(ii) Suppose x7 occurs at the (r + 1) th term in the given expression.

Then, we have:
Tr+1=Cr40 x40-r-1x2r =(-1)r Cr40 x40-r-2rFor this term to contain x7, we must have:40-3r=73r=40-7=33r=11Coefficient of x7 = (-1)11 C1140=-C1140

(iii) Suppose x−15 occurs at the (r + 1)th term in the given expression.
Then, we have:
Tr+1=Cr10 (3x2)10-r-a3x3r
Tr+1=(-1)r Cr10 310-r-rx20-2r-3rar
For this term to contain x-15 , we must have:20-5r =-155r=20+15r=7Coefficient of x-15 = (-1)7 C710 310-14 a7=-10×9×83×2×9×9a7=-4027a7

(iv) Suppose x9 occurs at the (r + 1)th term in the above expression.

Then, we have:

Tr+1=Cr9 (x2)9-r -13xr =(-1)r Cr9 x18-2r-r 13rFor this term to contain x9, we must have:18-3r=93r=9r=3Coefficient of x9 =(-1)3 C39 133=-9×8×72×9×9=-289

(v)
Suppose xm occurs at the (r + 1)th term in the given expression.

Then, we have:

Tr+1=Crn xn-r 1xr=Crn xn-2rFor this term to contain xm, we must have:n-2r =mr=(n-m)/2Coefficient of xm = C(n-m)/2n=n!n-m2! n+m2!

(vi) Suppose x occurs at the (r + 1)th term in the given expression.

Then, we have:

(1-2x3+3x5)1+1x8=1-2x3+3x5C08+C18 1x+C28 1x2+C38 1x3+C48 1x4+C58 1x5+C68 1x6+C78 1x7+C88 1x8 x occurs in the above expresssion at -2x3.C28 1x2 +3x5.C48 1x4.Coefficient of x =-28!2! 6!+38!4! 4!=-56+210= 154

(vii)
Suppose a5 b7 occurs at the (r + 1)th term in the given expression.

Then, we have:

Tr+1=Cr12 a12-r (-2b)r=(-1)r Cr12 a12-r br 2rFor this term to contain a5 b7, we must have:12-r =5 r=7 Required coefficient = (-1)7 C712 27=-12×11×10×9×8×1285×4×3×2=-101376

(viii) Suppose x occurs at the (r + 1)th term in the given expression.

Then, we have:

1-3x+7x2 1-x16=1-3x+7x2C016+C116 -x+C216 -x2+C316 -x3+C416 -x4+16C5 -x5+C616 -x6+C716 -x7+C816 -x8+C916 -x9+C1016 -x10+C1116 -x11+C1216 -x12+C1316 -x13+C1416 -x14+C1516 -x15+C1616 -x16 x occurs in the above expresssion at 16C1 -x -3xC016.Coefficient of x =-16!1! 15!-316!0! 16!=-16-3= -19

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integration of Irrational Algebraic Fractions - 1
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon