# Class 11 RD Sharma Solutions – Chapter 18 Binomial Theorem- Exercise 18.2 | Set 2

**Question 14. Find the middle terms in the expansion of:**

**(i) (3x â€“ x**^{3}/6)^{9}

^{3}/6)

^{9}

**Solution:**

We have,

(3x â€“ x

^{3}/6)^{9}where, n = 9 (odd number).So, the middle terms are ((n + 1)/2) = ((9 + 1)/2) = 10/2 = 5 and

((n + 1)/2 + 1) = ((9 + 1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6

The terms are 5

^{th}and 6^{th}.Now,

T

_{5}= T_{4+1}=

^{9}C_{4}(3x)^{9-4}(x^{3}/6)^{4}=

=

And, T

_{6}= T_{5+1}=

^{9}C_{5}(3x)^{9-5}(x^{3}/6)^{5}=

=

**(ii) (2x**^{2} â€“ 1/x)^{7}

^{2}â€“ 1/x)

^{7}

**Solution:**

We have,

(2x

^{2}â€“ 1/x)^{7}where, n = 7 (odd number).So the middle terms are ((n + 1)/2) = ((7 + 1)/2) = 8/2 = 4 and

((n + 1)/2 + 1) = ((7 + 1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5

The terms are 4

^{th}and 5^{th}.Now,

T

_{4}= T_{3+1}=

^{7}C_{3}(2x^{2})^{7-3}(-1/x)^{3}= -\frac{7Ã—6Ã—5}{3Ã—2}Ã—16x^8Ã—\frac{1}{x^3}

= âˆ’ 560 x

^{5}And, T

_{5}= T_{4+1}=

^{7}C_{4 }(2x^{2})^{7-4}(-1/x)^{4}=

= 280 x

^{2}

**(iii) (3x â€“ 2/x**^{2})^{15}

^{2})

^{15}

**Solution:**

We have,

(3x â€“ 2/x

^{2})^{15}where, n = 15 (odd number)So the middle terms are ((n + 1)/2) = ((15 + 1)/2) = 16/2 = 8 and

((n + 1)/2 + 1) = ((15 + 1)/2 + 1) = (16/2 + 1) = (8 + 1) = 9

The terms are 8

^{th}and 9^{th}.Now,

T

_{8}= T_{7+1}=

^{15}C_{7}(3x)^{15-7}(â€“ 2/x^{2})^{7}=

=

And, T

_{9}= T_{8+1}=

^{15}C_{8}(3x)^{15-8}(â€“ 2/x^{2})^{8}=

=

### (iv) (x^{4} â€“ 1/x^{3})^{11}

**Solution:**

We have,(x^{4}â€“ 1/x^{3})^{11 }where, n = 11 (odd number)

So the middle terms are ((n + 1)/2) = ((11 + 1)/2) = 12/2 = 6 and

((n + 1)/2 + 1) = ((11 + 1)/2 + 1) = (12/2 + 1) = (6 + 1) = 7The terms are 6

^{th}and 7^{th}.Now,

T

_{6}= T_{5+1}=

^{11}C_{5}(x^{4})^{11-5}(1/x^{3})^{5}=

= -462 x

^{9}And, T

_{7}= T_{6+1}=

^{11}C_{6}(x^{4})^{11-6}(1/x^{3})^{6}=

= 462 x

^{2}

**Question 15. Find the middle terms in the expansion of:**

**(i) (x â€“ 1/x)**^{10}

^{10}

**Solution:**

We have,

(x â€“ 1/x)

^{10 }where, n = 10 (even number)So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6

^{th}termNow,

T

_{6}= T_{5+1}=

^{10}C_{5}(x)^{10-5}(â€“1/x)^{5}=

= âˆ’252

### (ii) (1 – 2x + x^{2})^{n}

**Solution:**

We have, (1 – 2x + x

^{2})^{n}= (1 – x)

^{2n}Here, n is an even number.

So the middle terms is2n/2 + 1 = (n + 1)^{th }termNow,

T

_{n+1}=^{2n}C_{n}(-1)^{n}x^{n}=

### (iii) (1 + 3x + 3 x^{2} + x^{3})^{2n}

**Solution:**

We have, (1 + 3x + 3 x

^{2}+ x^{3})^{2n}= (1 + x )

^{6n}Here, n is an even number.

So the middle terms is(6n/2 + 1) = (3n + 1)^{th}termNow,

T

_{3n+1}=^{6n}C_{3n}x^{3n}=

**(iv) (2x â€“ x**^{2}/4)^{9}

^{2}/4)

^{9}

**Solution:**

We have,

(2x â€“ x

^{2}/4)^{9}where, n = 9 (odd number)So the middle terms are ((n + 1)/2) = ((9 + 1)/2) = 10/2 = 5 and

((n + 1)/2 + 1) = ((9 + 1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6

The terms are 5

^{th}and 6^{th}.Now,

T

_{5}= T_{4+1}=

^{9}C_{4}(2x)^{9-4}(â€“x^{2}/4)^{4}=

=

And, T

_{6}= T_{5+1}=

^{9}C_{5}(2x)^{9-5}(â€“x^{2}/4)^{5}=

=

### (v) (x – 1/x)^{2n+1}

**Solution:**

We have, (x – 1/x)

^{2n+1}Here, 2n + 1 is an odd number.

So the middle terms are((2n + 1 + 1)/2)^{th}and ((2n + 1 + 1)/2 + 1)^{th}terms.The terms are (n + 1)

^{th}and (n + 2)^{th}Now

T

_{n+1}==(-1)

^{n}^{2n+1}C_{n }xAnd T

_{n+2}= T_{n+1+1}=

=

### (vi) (x/3 + 9y)^{10}

**Solution:**

We have,

(x/3 + 9y)

^{10 }where, n = 10 (even number)So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6

^{th}term.Now,

T

_{6}= T_{5+1}=

^{10}C_{5}(x/3)^{10-5}(9y)^{5}=

= 61236 x

^{5}y^{5}

### (vii) (3 â€“ x^{3}/6)^{7}

**Solution:**

We have,

(3 â€“ x

^{3}/6)^{7}where, n = 7 (odd number).So the middle terms are ((n + 1)/2) = ((7 + 1)/2) = 8/2 = 4 and

((n + 1)/2 + 1) = ((7 + 1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5

The terms are 4

^{th}and 5^{th}.Now,

T

_{4}= T_{3+1}=

^{7}C_{3}(3)^{7-3}(-x^{3}/6)^{3}=

=

And, T_{5}= T_{4+1}

=^{9}C_{4}(3)^{9-4}(-x^{3}/6)^{4}

=

=

### (viii) (2ax â€“ b/x^{2})^{12}

**Solution:**

We have,

(2ax â€“ b/x

^{2})^{12}where, n = 12 (even number).So the middle terms are (n/2 + 1) = (12/2 + 1) = 7

^{th}termNow,

T

_{7}= T_{6+1}=

^{12}C_{6}(2ax)^{12-6}(-b/x^{2})^{6}=

^{12}C_{6}(2ax)^{6}(b/x^{2})^{6}=

^{12}C_{6}(2^{6}a^{6}x^{6})(b^{6}/x^{12})=

^{12}C_{6}(2^{6}a^{6}b^{6}/x^{6})

### (ix) (p/x + x/p)^{9}

**Solution:**

We have,

(p/x + x/p)

^{9}where, n = 9 (odd number).So the middle terms are ((n + 1)/2) = ((9 + 1)/2) = 10/2 = 5 and

((n + 1)/2 + 1) = ((9 + 1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6

The terms are 5

^{th}and 6^{th}.Now,

T

_{5}= T_{4+1}=

^{9}C_{4}(p/x)^{9-4}(x^{3}/p)^{4}=

^{9}C_{4}(p/x)^{5}(x/p)^{4}=

^{9}C_{4}(p/x)

And, T_{6}= T_{5+1}

=^{9}C_{5}(p/x)^{9-5}(x/p)^{5}

=^{9}C_{5 }(p/x)^{4}(x/p)^{5}

=^{9}C_{5 }(x/p)

### (x) (x/a â€“ a/x)^{10}

**Solution:**

We have,

(x/a â€“ a/x)

^{10}where, n = 10 (even number).So the middle terms are (n/2 + 1) = (10/2 + 1) 6

^{th}termNow,

T

_{6}= T_{5+1}=

^{10}C_{5}(x/a)^{10-5}(-a/x)^{5}= –

^{10}C_{5}(x/a)^{5}(a/x)^{5}= –

^{10}C_{5}= -252

**Question 16. Find the term independent of x in the expansion of the following expressions:**

**(i) (3/2 x**^{2} â€“ 1/3x)^{9}

^{2}â€“ 1/3x)

^{9}

**Solution:**

We have,

(3/2 x

^{2}â€“ 1/3x)^{9}We know, the (r+1)

^{th}term of the expression is given by,T

_{r+1}=^{ n}C_{r}x^{n-r}a^{r}=

^{9}C_{r}(3/2x^{2})^{9-r}(-1/3x)^{r}=

For this term to be independent of x, we must have

=> 18 â€“ 3r = 0

=> 3r = 18

=> r = 18/3

=> r = 6

So, the required term is 7

^{th}term.So, T

_{7}= T_{6+1}=

^{9}C_{6}Ã— (3^{9-12})/(2^{9-6})=

= 7/18

Hence, the term independent of x is 7/18.

**(ii) (2x + 1/3x**^{2})^{9}

^{2})

^{9}

**Solution:**

We have,

(2x + 1/3x

^{2})^{9}We know, the (r+1)

^{th}term of the expression is given by,T

_{r+1}=^{n}C_{r}x^{n-r}a^{r}=

^{9}C_{r}(2x)^{9-r}(1/3x^{2})^{r}=

For this term to be independent of x, we must have

=> 9 â€“ 3r = 0

=> r = 3

So, the required term is 4

^{th}term.So, T

_{7}= T_{6+1}=

^{9}C_{6}Ã— (2^{9-3})/(3^{3})=

= 5376/27

**(iii) (2x**^{2} â€“ 3/x^{3})^{25}

^{2}â€“ 3/x

^{3})

^{25}

**Solution:**

We have,

(2x

^{2}â€“ 3/x^{3})^{25}We know, the (r+1)

^{th}term of the expression is given by,T

_{r+1}=^{ n}C_{r}x^{n-r}a^{r}=

^{25}C_{r}(2x^{2})^{25-r}(-3/x^{3})^{r}= (-1)

^{r}^{25}C_{r}Ã— 2^{25-r}Ã— 3^{r}x^{50-2r-3r}For this term to be independent of x, we must have

=> 50 â€“ 5r = 0

=> 5r = 50

=> r = 10

So, the required term is 11

^{th}term.So, T

_{11 }= T_{10+1}= (-1)

^{10}^{25}C_{10}Ã— 2^{25-10}Ã— 3^{10}=

^{25}C_{10}(2^{15 }Ã— 3^{10})

**(iv) (3x â€“ 2/x**^{2})^{15}

^{2})

^{15}

**Solution:**

We have,

(3x â€“ 2/x

^{2})^{15}We know, the (r+1)

^{th}term of the expression is given by,T

_{r+1}=^{n}C_{r}x^{n-r}a^{r}=

^{15}C_{r}(3x)^{15-r}(-2/x^{2})^{r}= (-1)

^{r}^{15}C_{r}Ã— 3^{15-r}Ã— 2^{r}x^{15-r-2r}For this term to be independent of x, we must have

=> 15 â€“ 3r = 0

=> 3r = 15

=> r = 5

So, the required term is 6

^{th}term.So, T

_{6 }= T_{5+1}= (-1)

^{5}^{15}C5 Ã— 3^{15-5}Ã— 25= âˆ’3003 Ã— 3

^{10}Ã— 25

### (v)

**Solution:**

We have

We know, the (r+1)

^{th}term of the expression is given by,Now,

T

_{r+1}==

For this term to be independent of x, we must have

=> (10-r)/2 – 2r = 0

=> 10 – 5r = 0

=> r = 2

So, the required term is 3

^{th}term.So, T

_{3 }= T_{2+1}T

_{3}==

= 5/4

### (vi)

**Solution:**

We have

We know, the (r+1)

^{th}term of the expression is given by,Now,

T

_{r+1}== (-1)

^{r}^{3n}C_{r}x^{3n-r-2r}a^{r}For this term to be independent of x, we must have}

=> 3n – 3r = 0

=> r = n

So, the required term is (n + 1)

^{th}term.So,

T

_{n+1}= (-1)^{n 3n}C_{n}

### (vii)

**Solution:**

We have

We know, the (r+1)

^{th}term of the expression is given by,Now,

T

_{r+1}==

For this term to be independent of x, we must have

=> (8 – r)/3 – r/5 = 0

=> 40 – 5r – 3r = 0

=> 8r = 40

=> r = 5

So, the required term is 6

^{th}term.So, T

_{6 }= T_{5+1}T

_{6}==

= 7

### (viii) (1 + x + 2x^{3}) (3/2x^{2} â€“ 3/3x)^{9}

**Solution:**

We have

(1 + x + 2x

^{3}) (3/2x^{2}â€“ 3/3x)^{9}We know, the (r+1)

^{th}term of the expression is given by,Now,

T

_{r+1 }= (1 + x + 2x^{3}) [(3/2x^{2}) –^{9}C_{1}(3/2x^{2})^{8}(1/3x) + . . . –^{9}C_{7}(3/2x^{2})^{2}(1/3x)^{7}]For this term to be independent of r, we must have

=

^{9}C_{6}(3^{3}/2^{3}) (1/3^{6}) – 2x^{3}^{9}C_{7}(2^{3}/3^{3}) (1/3^{7}) (1/x^{3})= 7/18 â€“ 2/27

= (189 â€“ 36)/486

= 153/486

= 17/54

### (ix) , x > 0

**Solution:**

We have

We know, the (r+1)

^{th}term of the expression is given by,Now,

T

_{r+1}==

For this term to be independent of r, we must have

=> (18 – r)/3 – r/3 = 0

=> 18 – 2r = 0

=> r = 9

So, the required term is 9

^{th}term.So, T

_{9 }= T_{8+1}T

_{9}=^{18}C_{9}(1/2^{9})

### (x)

**Solution:**

We have

Suppose the (r + 1)

^{th}term in the given expression is independent of x.Now,

T

_{r+1}==

For this term to be independent of x, we must have

=> 12 – 3r = 0

=> r = 4

Hence, the required term is the 4

^{th}term.So, T

_{4 }= T_{3+1}=

= 5/12

**Question 17. If the coefficients of (2r + 4)**^{th} and (r â€“ 2)^{th} terms in the expansion of (1 + x)^{18} are equal, find r.

^{th}and (r â€“ 2)

^{th}terms in the expansion of (1 + x)

^{18}are equal, find r.

**Solution:**

We are given,

(1 + x)

^{18}We know, the coefficient of the r

^{th}term in the expansion of (1 + x)^{n}is^{n}C_{r-1}.So, the coefficients of the (2r + 4)

^{th}and (r â€“ 2)^{th}terms in the given expansion are,

^{18}C_{2r+4-1}and^{18}C_{r-2-1}According to the question, we have,

=>

^{18}C_{2r+4-1}=^{18}C_{r-2-1}=>

^{18}C_{2r+3}=^{18}C_{r-3}We know if

^{n}C_{r }=^{n}C_{s}, then r = s or r + s = n.=> 2r + 3 = r â€“ 3 or 2r + 3 + r â€“ 3 = 18

=> 2r â€“ r = â€“3 â€“ 3 or 3r = 18 â€“ 3 + 3

=> r = â€“6 or 3r = 18

=> r = â€“6 or r = 6

Ignoring r = â€“ 6 as r cannot be negative.

Therefore, the value of r is 6.

**Question 18. If the coefficients of (2r + 1)**^{th} term and (r + 2)^{th} term in the expansion of (1 + x)^{43} are equal, find r.

^{th}term and (r + 2)

^{th}term in the expansion of (1 + x)

^{43}are equal, find r.

**Solution:**

We are given,

(1 + x)

^{43}We know, the coefficient of the r

^{th}term in the expansion of (1 + x)^{n}is^{n}C_{r-1}.So, the coefficients of the (2r + 1)

^{th}and (r + 2)^{th}terms in the given expansion are,

^{43}C_{2r+1-1}and^{43}C_{r+2-1}According to the question, we have,

=>

^{43}C_{2r+1-1}=^{43}C_{r+2-1}=>

^{43}C_{2r }=^{43}C_{r+1}We know if

^{n}C_{r}=^{n}C_{s}, then r = s or r + s = n.=> 2r = r + 1 or 2r + r + 1 = 43

=> r = 1 or 3r = 42

=> r = 1 or r = 14

Ignoring r = 1 as it gives the same term on both the sides.

Therefore, the value of r is 14.

**Question 19. Prove that the coefficient of (r + 1)**^{th} term in the expansion of (1 + x)^{n+1} is equal to the sum of the coefficients of r^{th} and (r+1)^{th} terms in the expansion of (1 + x)^{n}.

^{th}term in the expansion of (1 + x)

^{n+1}is equal to the sum of the coefficients of r

^{th}and (r+1)

^{th}terms in the expansion of (1 + x)

^{n}.

**Solution:**

We know, the coefficients of (r + 1)

^{th}term in (1 + x)^{n+1}is^{n+1}C_{r}.So, sum of the coefficients of the r

^{th}and (r + 1)^{th }terms in (1 + x)^{n}is,(1 + x)

^{n}=^{n}C_{r-1}+^{n}C_{r}As,

^{n}C_{r+1}+^{n}C_{r}=^{n+1}C_{r+1}=

^{n+1}C_{r}

Hence proved.

**Question 20. Prove that the term independent of x in the expansion of (x + 1/x)**^{2n} is **.**

^{2n}is

**Solution:**

We have,

(x + 1/x)

^{2n}We know the (r + 1)

^{th}term is given by,T

_{r+1}=^{ n}C_{r}x^{n-r}a^{r}=

^{2n}C_{r }x^{2n-r}(1/x)^{r}=

^{2n}C_{r}x^{2n-2r}For this term to be independent of x, we must have,

=> 2n âˆ’ 2r = 0

=> 2n = 2r

=> r = n

Therefore, the required term is (n+1)

^{th}term .T

_{n+1}=^{2n}C_{n}x^{2n-n}(1/x)^{n}=

^{2n}C_{n}=

=

=

=

=

Hence proved.

**Question 21. If the coefficients of 5**^{th}, 6^{th} and 7^{th} terms of the expansion (1 + x)^{n} are in A.P., find n.

^{th}, 6

^{th}and 7

^{th}terms of the expansion (1 + x)

^{n}are in A.P., find n.

**Solution:**

We have, (1 + x)

^{n}We know the coefficient of r

^{th}term of a binomial expression is given by^{n}C_{r-1}.Coefficient of 5

^{th}term =^{n}C_{5-1}=^{n}C_{4}Coefficient of 6

^{th}term =^{n}C_{6-1}=^{n}C_{5}Coefficient of 7

^{th}term =^{n}C_{7-1}=^{n}C_{6}According to the question, we have,

=> 2

^{n}C_{5}=^{n}C_{4}+^{n}C_{6}=>

=>

=>

=>

=>

=>

=> 60(nâˆ’4) = 150 + 5n

^{2 }âˆ’ 45n + 100=> 60n âˆ’ 240 = 250 + 5n

^{2}âˆ’ 45n=> 5n

^{2}âˆ’ 105n + 490 = 0=> n

^{2}âˆ’ 21n + 98 = 0=> n

^{2}âˆ’ 7n âˆ’ 14n + 98 = 0=> n (n âˆ’ 7) âˆ’ 14 (n âˆ’ 7) = 0

=> (n âˆ’ 7) (n âˆ’ 14) = 0

=> n = 7 or n = 14

Therefore, the value of n is 7 or 14.

**Question 22. If the coefficients of 2nd, 3rd**,** and 4th terms of the expansion (1 + x)**^{2n} are in A.P., show that 2n^{2} âˆ’ 9n + 7 = 0.

^{2n}are in A.P., show that 2n

^{2}âˆ’ 9n + 7 = 0.

**Solution:**

We have, (1 + x)

^{2n}We know the coefficient of r

^{th}term of a binomial expression is given by^{n}C_{r-1}.Coefficient of 2

^{nd}term =^{2n}C_{2-1}=^{2n}C_{1}Coefficient of 3

^{rd}term =^{2n}C_{3-1}=^{2n}C_{2}Coefficient of 4

^{th}term =^{2n}C_{4-1}=^{2n}C_{3}According to the question, we have,

=> 2

^{2n}C_{2}=^{2n}C_{1}+^{2n}C_{3}=> 2 =

=> 2 =

=> = 2

=> = 2

=> 4n

^{2}âˆ’ 6n + 8 = 12n âˆ’ 6=> 4n

^{2}âˆ’ 18n + 14 = 0=> 2 (2n

^{2}âˆ’ 9n + 7) = 0=> 2n

^{2}âˆ’ 9n + 7 = 0

Hence proved.

**Question 23. In the expansion of (1 + x)**^{n}, the binomial coefficients of three consecutive terms are respectively 220, 495,** and 792, find the value of n.**

^{n}, the binomial coefficients of three consecutive terms are respectively 220, 495

**Solution:**

We have, (1 + x)

^{n}Let the three consecutive terms be r

^{th}, (r+1)^{th}and (r+2)^{th}.We know the coefficient of r

^{th}term of a binomial expression is given by^{n}C_{r-1}.Coefficient of r

^{th}term =^{n}C_{r-1}= 220Coefficient of (r+1)

^{th}term =^{n}C_{r+1-1}=^{n}C_{r}= 495Coefficient of (r+2)

^{th}term =^{n}C_{r+2-1}=^{n}C_{r+1}= 792Now,

=>

=>

=> 5n âˆ’ 5r = 8r + 8

=> 5n âˆ’ 13r = 8 . . . . (1)

Also,

=>

=> 4n âˆ’ 4r + 4 = 9r

=> 4n âˆ’ 13r = âˆ’4 . . . . (2)

Subtracting (2) from (1), we get,

=> n = 8 + 4

=> n = 12

Therefore, the value of n is 12.

**Question 24. If the coefficients of 2**^{nd}, 3^{rd} and 4^{th} terms of the expansion (1 + x)^{n} are in A.P., then find the value of n.

^{nd}, 3

^{rd}and 4

^{th}terms of the expansion (1 + x)

^{n}are in A.P., then find the value of n.

**Solution:**

We have, (1 + x)

^{n}We know the coefficient of r

^{th}term of a binomial expression is given by^{n}C_{r-1}.Coefficient of 2

^{nd}term =^{n}C_{2-1}=^{n}C_{1}Coefficient of 3

^{rd}term =^{n}C_{3-1}=^{ n}C_{2}Coefficient of 4

^{th}term =^{n}C_{4-1}=^{n}C_{3}According to the question, we have,

=> 2

^{n}C_{2}=^{n}C_{1}+^{n}C_{3}=>

=>

=>

=> = 2

=> n

^{2}âˆ’ 3n + 8 = 6 (n âˆ’ 1)=> n

^{2}âˆ’ 3n + 8 = 6n âˆ’ 6=> n

^{2}âˆ’ 9n + 14 = 0=> n

^{2}âˆ’ 7n âˆ’ 2n + 14 = 0=> n (nâˆ’7) âˆ’ 2 (nâˆ’7) = 0

=> (n âˆ’ 2) (n âˆ’ 7) = 0

=> n = 2 or n = 7

Ignoring n = 2 as it does not satisfy our condition.

Therefore, the value of n is 7.

**Question 25. If in the expansion of (1 + x)**^{n}, the coefficients of p^{th} and q^{th} terms are equal, then prove that p + q = n + 2, where p â‰ q.

^{n}, the coefficients of p

^{th}and q

^{th}terms are equal, then prove that p + q = n + 2, where p â‰ q.

**Solution:**

We have, (1 + x)

^{n}We know the coefficient of r

^{th}term of a binomial expression is given by^{n}C_{r-1}.Coefficient of p

^{th}term =^{n}C_{p-1}Coefficient of q

^{th}term =^{n}C_{q-1}According to the question, we have,

=>

^{n}C_{p-1}=^{n}C_{q-1}=> p âˆ’ 1 = q âˆ’ 1 or p âˆ’ 1 + q âˆ’ 1 = n

=> p = q or p + q âˆ’ 2 = n

=> p + q âˆ’ 2 = n

=> p + q = n + 2

Hence proved.

**Question 26. If in the expansion of (1 + x)**^{n}, the binomial coefficients of three consecutive terms are respectively 56, 70,** and 56, find n and the position of the terms of these coefficients.**

^{n}, the binomial coefficients of three consecutive terms are respectively 56, 70

**Solution:**

We have, (1 + x)

^{n}Let the three consecutive terms be r

^{th}, (r+1)^{th}and (r+2)^{th}.We know the coefficient of r

^{th}term of a binomial expression is given by^{n}C_{r-1}.Coefficient of r

^{th}term =^{n}C_{r-1}= 56Coefficient of (r+1)

^{th}term =^{n}C_{r+1-1}=^{n}C_{r}= 70Coefficient of (r+2)

^{th}term =^{n}C_{r+2-1}=^{ n}C_{r+1}= 56Now,

=>

=>

=> 5n âˆ’ 5r = 4r + 4

=> 5n âˆ’ 9r = 4 . . . . (1)

Also,

=>

=> 4n âˆ’ 4r + 4 = 5r

=> 4n âˆ’ r = âˆ’4 . . . . (2)

Subtracting (2) from (1), we get,

=> n = 4 + 4

=> n = 8

Putting n = 8 in (1), we get,

=> 5(8) âˆ’ 9r = 4

=> 40 âˆ’ 9r = 4

=> 9r = 36

=> r = 4

Therefore, three consecutive terms are 4^{th}, 5^{th}and 6^{th}terms.

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