Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+4 votes
23.9k views
in Mathematics by (63.2k points)

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

1 Answer

+3 votes
by (13.2k points)
selected by
 
Best answer

Solution: Given the coefficients of( r-1)th, rth and (r+1)th terms in the expansion of (x+1)n are in the ratio 1:3:5

Coefficient of (r-1)th term = C(n, r-2)

Coefficient of rth term = C(n, r-1)

Coefficient of (r+1)th term = C(n, r)

C(n, r-2) : C(n, r-1) : C(n, r) = 1:3:5

Divide 1st and 2nd

C(n,r-2)/C(n,r-1) = 1 / 3

= [n! / (r-2)! (n -r +2)! ] x [(r-1)!(n-r+1)!/ n!] = 1/3


( r -1)(r -2)! / ( r – 2)! ( n -r +2) = 1 / 3

( r -1) / ( n -r +2) = 1 / 3

3r - 3 = n - r + 2

n - 4r = -5 ----------(1)

Divide 2nd and 3rd

C(n,r-1)/C(n,r) = 3/5 

[n! / (r-1)! (n -r +1)! ] x r!(n-r)!/ n! = 3 / 5

r / (n -r + 1) = 3 / 5
 
5r = 3n - 3r +3

3n - 8r = -3 ---------(2)

2(n - 4r = -5)

2n - 8r = -10 ---------(3) 

Subtract (3) from (2)

n = 7

Substitute n = 7 in (2)

We get r = 3

n = 7, r = 3

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...