Q. 103.7( 70 Votes )

# The coefficients of the (r – 1)^{th}, r^{th} and (r + 1)^{th} terms in the expansion of (x + 1)^{n} are in the ratio 1 : 3 : 5. Find n and r.

Answer :

The general term T_{r+1} in the binomial expansion is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Here the binomial is (1+x)^{n} with a = 1 , b = x and n = n

The (r+1)^{th} term is given by

T_{(r+1)} = ^{n}C_{r} 1^{n-r} x^{r}

T_{(r+1)} = ^{n}C_{r} x^{r}

The coefficient of (r+1)^{th} term is ^{n}C_{r}

The r^{th} term is given by (r-1)^{th} term

T_{(r+1-1)} = ^{n}C_{r-1} x^{r-1}

T_{r} = ^{n}C_{r-1} x^{r-1}

∴ the coefficient of r^{th} term is ^{n}C_{r-1}

For (r-1)^{th} term we will take (r-2)^{th} term

T_{r-2+1} = ^{n}C_{r-2} x^{r-2}

T_{r-1} = ^{n}C_{r-2} x^{r-2}

∴ the coefficient of (r-1)^{th} term is ^{n}C_{r-2}

Given that the coefficient of (r-1)^{th}, r^{th} and r+1^{th} term are in ratio

1:3:5

⇒ 3r - 3 = n – r + 2

⇒ n - 4r + 5 =0…………1

Also

⇒

⇒

⇒

⇒

⇒

⇒

⇒ 5r = 3n - 3r + 3

⇒ 8r – 3n - 3 =0………….2

We have 1 and 2 as

n - 4r 5 =0…………1

8r – 3n - 3 =0…………….2

Multiplying equation 1 by number 2

2n -8r +10 =0……………….3

Adding equation 2 and 3

2n -8r +10 =0

+ -3n – 8r - 3 =0

⇒ -n = -7

**n =7 ****and r = 3**

Rate this question :

Find the coefficient of

a^{5}b^{7} in (a – 2b)^{12} .

Show that the coefficient of x^{4} in the expansion of is .

Show that the middle term in the expansion of is 252.

RS Aggarwal - MathematicsShow that the coefficient of x^{-3} in the expansion of is -330.

Show that the term independent of x in the expansion of is -252.

RS Aggarwal - MathematicsFind the middle term in the expansion of

RS Aggarwal - MathematicsIf the 17^{th} and 18^{th} terms in the expansion of (2 + a)^{50} are equal, find the value of a.