﻿ Class 11 NCERT Math Solution
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TOPICS
Unit-8(Examples)

Example-1 :-  Expand (x2 + 3/x)4, x ≠ 0

Solution :-
```  (x2 + 3/x)4
= 4 C0(x2)4 + 4C1(x2)3(3/x) + 4C2(x2)2(3/x)2 + 4C3(x2)(3/x)3 + 4C4(3/x)4
= x8 + 4x6 . 3/x + 6x4 . 9/x2 + 4x2 . 27/x3 + 81/x4
= x8 + 12x5 + 54x2 + 108/x + 81/x4
```

Example-2 :-  Compute (98)5.

Solution :-
```  985 = (100 - 2)5
= 5C0(100)5 - 5C1(100)4.2 + 5C2(100)3.22 - 5C3(100)2.23 + 5C4(100).24 - 5C5(2)5
= 100000000000 - 5 x 100000000 x 2 + 10 x 1000000 x 4 - 10 x 10000 x 8 + 5 x 100 x 16 - 32
= 10040008000 - 1000800032
= 9039207968
```

Example-3 :-  Which is larger (1.01)1000000 or 10,000?

Solution :-
```  (1.01)1000000 = (1 + 0.01)1000000
= 1000000C0 + 1000000C1(0.01) + other positive terms
= 1 + 1000000 × 0.01 + other positive terms
= 1 + 10000 + other positive terms
> 10000
Hence (1.01)1000000 > 10000
```

Example-4 :-  Using binomial theorem, prove that 6n – 5n always leaves remainder 1 when divided by 25.

Solution :-
```  For two numbers a and b if we can find numbers q and r such that a = bq + r,
then we say that b divides a with q as quotient and r as remainder.
Thus, in order to show that  6n – 5n leaves remainder 1 when divided by 25,
we prove that 6n – 5n = 25k + 1, where k is some natural number.
We have (1 + a)n = nC0 + nC1a + nC2a2 + ... + nCnan
For a = 5, we get
(1 + 5)n = nC0 + nC15 + nC252 + ... + nCn5n i.e.
(6)n = 1 + 5n + 52.nC2 + 53.nC3 + ... + 5n i.e.
6n – 5n = 1 + 52(nC2 + nC35 + ... + 5n-2) or
6n – 5n = 1 + 25(nC2 + 5 .nC3 + ... + 5n-2) or
6n – 5n = 25k + 1
where k =  nC2 + 5 .nC3 + ... + 5n-2.
This shows that when divided by 25, 6n – 5n leaves remainder 1.
```

Example-5 :-  Find a if the 17th and 18th terms of the expansion (2 + a)50 are equal.

Solution :-
```  The (r + 1)th term of the expansion (x + y)n is given by Tr+1 =  nCr xn-r yr.
For the 17th term, we have, r + 1 = 17, i.e., r = 16
Therefore, T17 = T16+1 = 50C16 (2)50–16 a16 = 50C16 234 a16.
Similarly, T18 = 50C17 233 a17
Given that T17 = T18
So, 50C16 (2)34 a16 = 50C17 (2)33 a17
Therefore,
[50C16 (2)34] ÷ [50C17 (2)33] = a17 ÷ a16
a = 50!/(16!34!) x (17!33!)/50! x 2 = 1
```

Example-6 :-  Show that the middle term in the expansion of (1+x)2n is [1.3.5...(2n-1)]/n! 2n xn, where n is a positive integer.

Solution :-
```  As 2n is even, the middle term of the expansion (1 + x)2n is (2n/2 + 1)th,
i.e., (n + 1)th term which is given by,

```

Example-7 :-  Find the coefficient of x6y3 in the expansion of (x + 2y)9.

Solution :-
```  Suppose x6y3 occurs in the (r + 1)th term of the expansion (x + 2y)9.
Now Tr+1  =  9Cr x9-r (2y)r  =  9Cr 2r . x9-r . yr .
Comparing the indices of x as well as y in x6y3 and in  Tr+1 , we get r = 3.
Thus, the coefficient of x6y3 is
9C3 23 = 9!/(3!6!) x 23 = 672
```

Example-8 :-  The second, third and fourth terms in the binomial expansion (x + a)n are 240, 720 and 1080, respectively. Find x, a and n.

Solution :-
```  Given that second term T2 = 240
We have T2 = nC1 xn-1 . a
So nC1 xn-1 . a = 240 ... (1)
Similarly nC2 xn-2 a2 = 720 ... (2)
and nC3 xn-3 a3 = 1080 ... (3)
Dividing (2) by (1), we get
[nC2 xn-2 a2] ÷ [nC1 xn-1 . a] = 720 ÷ 240
[(n-1)!.a] ÷ [(n-2)!.x] = 6   ....(4)
a/x = 6/(n-1)
Dividing (3) by (2), we have
[nC3 xn-3 a3] ÷ [nC2 xn-2 a2] = 1080 ÷ 720
a/x = 9/[2(n-2)] ....(5)
From (4) and (5),
6/(n-1) = 9/[2(n-2)]
9n - 9 = 12n - 24
12n - 9n = 24 - 9
3n = 15
n = 5
Hence, from (1), 5x4a = 240,
and from (4), a/x = 3/2
Solving these equations for a and x, we get x = 2 and a = 3.
```

Example-9 :-  The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio1: 7 : 42. Find n.

Solution :-
```  Suppose the three consecutive terms in the expansion of (1 + a)n are (r – 1)th, rth and (r + 1)th terms.
The (r – 1)th term is nCr-2 ar-2, and its  coefficient is nCr-2.
Similarly, the coefficients of rth and (r + 1)th terms  are nCr-1 and nCr, respectively.
Since the coefficients are in the ratio 1 : 7 : 42, so we have,
nCr-2 : nCr-1 = 1/7,  i.e., n – 8r + 9 = 0 ... (1)
nCr-1 : nCr = 7/42 , i.e., n – 7r + 1 = 0 ... (2)
Solving equations(1) and (2), we get, n = 55.
```
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