Find the term independent of x in the expasion of the following expressions:
(i)(32x2−13x)9(ii)(2x+13x2)9(iii)(2x2−33x3)25(iv)(3x−22x2)15(v)(√x3+32x210)(vi)(x−1x2)3n(vii)(12x1/3+x−1/5)8(viii)(1+x+2x3)(32x2−13x)9(ix)(3√x+123√x)18,x>2(x)(32x2−13x)6
(i)(32x2−13x)
Tr+1=9Cr(3x22)9−r(−13x)r
=9Cr(32)9−r(x18−2r)(−13)rx−r
Let T{r+1}~be~independent~of~x 18−3r=0 or r=6 ∴Required term
⇒Tr+1=T6+1=T7=9C6(32)9−6
=84(278)(1179)x0=718
(ii)(2x+13x2)
Suppose the (r+1th)term in the given expression is independent of x
Now (2x+13x2)94
Tr+1=9Cr(2x)9−r(13x2)r
=9Cr.29−r3rx9−r−2r
For this term to be independent of x, we must have 9 -3r = 0
⇒r=3
Hence, the required term is the 4th term.
Now, we have
9C32633=9C3×6427
(iii)(2x2−3x3)25
Tr+1=(−1)r nCr(2x2)25−r(3x3)r
=(−1)rnCr2x25−r3r x50−2r−3r
Term independent of x=x0
⇒x50−50r=x0
⇒50−5r=0⇒r=10
∴t11=(−1)1025C10215×310
=25C10215310
(iv)(3x−2x2)15
Tr+1=(−1)r 15Cr(3x)15−r(2x2)r
=(−1)r 15Cr315−r2r x15−r−1r
Term independent of x⇒x0
⇒x15−3r=x0
15−3r=0⇒r=5
∴t6=(−1)5 15C5 31025
=−15!5!10!×31025
=−15×14×13×12×11120 31025
=−3003×310×25
(V)(√x3+32x2)10
Tr+1=10Cr(√x3)10−r(32x2)r
=10Cr x5−r2−2r×3r×3−5+r2×2−r
Independent of x⇒x0
x10−r−4rr=x0
10−5r=0
r=2
t3−10C2 32−5+12−2
=10C23−22−2
=10!2!8!×136=10×92×36=54
(vi)(x−1x2)3n
Suppose the(r+1)thterm in the given expression is independent of x
Now, (x−1x2)3n
Tr+1=3nCr x3n−r (−1x2)r
=(−1)r 3nCr x3n−r−2r
For this term to be independent of x, we must have 3n - 3r = 0
⇒r=n Hence, the required term is the (n+1)th term.
Now, we have
(−1)n3nCn
(vii)(12x1/3+x−1/5)8
We have,
(12x1/3+x−1/5)8
Let(r+1)thterm be independent of x
∴Tr+1=8Cr(12x13)8−r(x−15)r
=8Cr(12)8−r×(x13)8−r×(1x15)r
=8Cr(12)8−r×(x)8−r3×(1x15)
=8Cr(12)8−r×(x)8−r r3 5
=8Cr(12)8−r×(x)40−5r−3r15
=8Cr(12)8−r×(x)40−8r15
It it is independent of x, we must have =40−8r15=0
⇒8r=40
⇒r=5
∴The term independent ofx=T6
Now
T6=8Cr(12x13)8−5(x−15)5
=56×(12)3
=56×18
=7 Hence, required term = 7.
(viii)(1+x+2x3)(32x2−13x)9
=(1+x+2x3)(32x2−13x)9
=(1+x+2x3)
⎡⎢ ⎢ ⎢⎣(32x2)9−9C1(32x2)813x...+9C6(32x2)3(13x6−9C7(32x2)2(13x7))⎤⎥ ⎥ ⎥⎦
In the second bracket, we have to search the term so x0 and 1x3 which when multiplying by 1 and 2x3 is firs bracket will give the term in dependent of x. The term containing 1x will not occur is second bracket. The term independent of x =[9C6 3323×136]−2x3[9C7 3323×137×1x3]
=[9×8×71×2×3×18×27]−2[9×81×2−14×243] =718−227
=1754
Required term=1754.
(ix)(3√x+123√x)18,x>2We have,
(3√x+13√x)18,x>0
Let(r+1)thterm be independent of x.
∴Tr+1=18Cr(3√x)18−r×(123√x)r
=18Cr((x)13)18−r×(12)r×(1x13)r =18Cr(x)18−r3×(1xr3)×(12)r
=18Cr(x)18−r3×(12)r
=18Cr(x)18−r3×(12)r
If it is independent of x, we must have 18−2r3=0
⇒18=2r ⇒r=9
∴Term independent ofx=T9+1=T10
Now, T10=18C9(3√x)18−9(123√x)9
=18C9(3√x)9×129×(13√x)9
=18C929
Hence, required term =18C929.
(x)(3xx2−13x)6
In expansion
Tr+1=6Cr(3x22)6−2(−13x)r
=6Cr(32)6−2(x12−3r)(−13)r
LetTr+1be independent of x,
12−3r−0orr=4
∴Required term
⇒Tr+1=T4+1=T5=6C4(32)6−4x12−3(4)
=15(94)(118)x∘=512