﻿ Class 11 NCERT Math Solution
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Exercise - 8.2

Question-1 :-  Find the coefficient of x5 in (x + 3)8 .

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr.
Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8,
we obtain Tr+1 = 8Cr x8-r 3r
Comparing the indices of x in x5 and in Tr+1, we obtain r = 3
Thus, the coefficient of x5 is 8C3 33 = 8!/3!5! x 33 = 1512
```

Question-2 :- Find the coefficient of a5b7 in (a – 2b)12.

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr.
Assuming that a5b7 occurs in the (r + 1)th term of the expansion (a – 2b)12,
we obtain Tr+1 = 12Cr a12-r (-2b)r = 12Cr (-2)r a12-r (b)r
Comparing the indices of a and b in a5 b7 and in Tr+1, we obtain r = 7
Thus, the coefficient of a5b7 is 12C7 (-2)7 = 12!/7!5! x (-2)7 = (-792)(128) = -101376
```

Question-3 :-  Write the general term in the expansion of (x2 – y)6.

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr.
Thus, the general term in the expansion of (x2 – y)6 is
Tr+1 = 6Cr (x2)6-r (-y)r = 6Cr (-1)r x12-2r yr
```

Question-4 :-  Write the general term in the expansion of (x2 – yx)12, x ≠ 0.

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr.
Thus, the general term in the expansion of (x2 – yx)12 is
Tr+1 = 12Cr (x2)12-r (-yx)r
= 12Cr (-1)r x24-2r xr yr
= 12Cr (-1)r x24-r yr
```

Question-5 :-  Find the 4th term in the expansion of (x – 2y)12.

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr.
Thus, the 4th term in the expansion of (x – 2y)12 is
T4 = T3+1
= 12C3 x12-3 (-2y)3
= (-1)3 12!/3!9! x9 8y3
= -1760 x9 y3
```

Question-6 :-  Find the 13th term in the expansion of (9x - 1/3√x)18, x ≠ 0.

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr.
Thus, the 13th term in the expansion of (9x – 1/3√x)18 is
T13 = T12+1
= 18C12 (9x)18-12 (-1/3√x)12
= (-1)12 18!/12!6! (9)6 (x)6 (1/3)12 (1/√x)12
= 18!/12!6! (3)12 (x)6 (1/3)12 (1/x)6
= 18!/12!6!
= 18564
```

Question-7 :-  Find the middle terms in the expansions of (3 - x3/6)7.

Solution :-
```  It is known that in the expansion of (a + b)n, if n is odd,
then there are two middle terms, namely, [(n+1)/2]th term and [(n+1)/2 + 1] term.
Therefore, the middle terms in the expansion of (3 - x3/6)7 are [(7+1)/2]th term and [(7+1)/2 + 1]th term.
Now, T4 = T3+1
= 7C3 37-3 (-x3/6)3
= (-1)3 7!/3!4! 34 (x9/63)
= -105x9/8
Again, T5 = T4+1
= 7C4 37-4 (-x3/6)4
= (-1)4 7!/3!4! 33 (x12/64)
= 35x12/48

Thus, the middle terms in the expansion of (3 - x3/6)7 are -105x9/8 and 35x12/48.
```

Question-8 :-  Find the middle terms in the expansions of (x/3 + 9y)10.

Solution :-
```  It is known that in the expansion of (a + b)n, if n is even,
then there are two middle terms, namely, [n/2 + 1]th term.
Therefore, the middle terms in the expansion of (x/3 + 9y)10 are [10/2 + 1]th = 6th term.
Now, T6 = T5+1
= 10C5 (x/3)10-5 (9y)5
= 95 10!/5!5! x5 (1/3)5 y5
= 310 10!/5!5! x5 1/35 y5
= 61236 x5 y5

Thus, the middle terms in the expansion of (x/3 + 9y)10 is 61236 x5 y5.
```

Question-9 :-  In the expansion of (1 + a)m+n, prove that coefficients of am and an are equal.

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr.
Assuming that am occurs in the (r + 1)th term of the expansion (1 + a)m+n,
we obtain Tr+1 = m+nCr 1m+n-r ar = m+nCr ar
Comparing the indices of a in am and in Tr+1, we obtain r = m
Therefore, the coefficient of am is
m+nCm = (m+n)!/[m!(m+n-m)!] = (m+n)!/m!n! .....(i)

Assuming that an occurs in the (k + 1)th term of the expansion (1 + a)m+n,
we obtain Tk+1 = m+nCk 1m+n-k ak = m+nCk ak
Comparing the indices of a in an and in Tk+1, we obtain k = n
Therefore, the coefficient of an is
m+nCn = (m+n)!/[n!(m+n-n)!] = (m+n)!/m!n! .....(ii)

Thus, from (i) and (ii), it can be observed that the coefficients of am and an in the expansion of (1 + a)m+n are equal.
```

Question-10 :-  The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r.

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr.
Therefore, (r – 1)th term in the expansion of (x + 1)n is Tr-1 = nCr-2 xn-(r-2) 1r-2 = nCr-2 xn-r+2
Now, rth term in the expansion of (x + 1)n is Tr = nCr-1 xn-(r-1) 1r-1 = nCr-1 xn-r+1
Again, (r + 1)th term in the expansion of (x + 1)n is Tr+1 = nCr xn-r 1r

Therefore, the coefficients of the (r – 1)th, rth, and (r + 1)th terms in the expansion of (x + 1)n are nCr-2, nCr-1 and nCr respectively.
Since these coefficients are in the ratio 1:3:5, we obtain
nCr-2 : nCr-1 = 1 : 3
n!/[(r-2)!(n-r+2)!] x [(r-1)!(n-r+1)!]/n! = 1/3
(r-1)/(n-r+2) = 1/3
n - r + 2 = 3r - 3
n - 4r + 5 = 0 .....(i)

nCr-1 : nCr = 3 : 5
n!/[(r-1)!(n-r+1)!] x [r!(n-r)!]/n! = 3/5
r/(n-r+1) = 3/5
3n - 3r + 3 = 5r
3n - 8r + 3 = 0 .....(ii)

Multiplying (i) by 3 and subtracting it from (ii), we obtain 4r – 12 = 0, r = 3
Putting the value of r in (i), we obtain n – 12 + 5 = 0, n = 7
Thus, n = 7 and r = 3
```

Question-11 :-  Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n-1.

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr.
Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n, we obtain Tr+1 = 2nCr (1)2n-r xr = 2nCr xr
Comparing the indices of x in xn and in Tr+1, we obtain r = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n is 2nCn = (2n)!/n!n! = (2n)!/(n!)2  .......(i)

Assuming that xn occurs in the (k + 1)th term of the expansion (1 + x)2n-1, we obtain Tk+1 = 2n-1Ck (1)2n-1-k xk = 2n-1Ck xk
Comparing the indices of x in xn and Tk+1, we obtain k = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n-1 is
2n-1Ck = (2n-1)!/[n!(n-1)!] = (2n)!/(2.n!n!) = 1/2 x [(2n)!/(n!)2] .....(ii)

From (i) and (ii), it is observed that
1/2 x 2nCn = 2n-1Cn
2nCn = 2.2n-1Cn
Therefore, the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n-1.
Hence, proved.
```

Question-12 :-  Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr.
Assuming that x2 occurs in the (r + 1)th term of the expansion (1 + x)m, we obtain Tr+1 = mCr (1)m-r xr = mCr xr
Comparing the indices of x in x2 and in Tr+1, we obtain r = 2
Therefore, the coefficient of x2 is mC2.
It is given that the coefficient of x2 in the expansion (1 + x)m is 6.
Therefore, mC2 = 6
m!/[2!(m-2)!] = 6
[m x (m-1) x (m-2)!]/[2 x 1 x (m-2)!] = 6
[m(m-1)]/2 = 6
m2 - m = 12
m2 - m - 12 = 0
m2 - 4m + 3m - 12 = 0
m(m-4) + 3(m-4) = 0
(m + 3)(m - 4) = 0
m = -3 or m = 4
Thus, the positive value of m, for which the coefficient of x2 in the expansion (1 + x)m is 6, is 4.
```
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