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
0 votes
4.4k views
in Mathematical Induction by (22.8k points)
closed by

If ‘n’ be any positive integer, then n (n + 1) (2n + 1) is

(a) an odd integer 

(b) an integral multiple of 6 

(c) a perfect square 

(d) None of these

1 Answer

+1 vote
by (23.7k points)
selected by
 
Best answer

Answer: (B) an integral multiple of 6 

When n = 1, n(n+1)(2n+1) = (1) (2) (3) = 6, which is an integral multiple of 6. 

It is neither an odd integer nor a perfect square. 

Using the principle of mathematical induction, we shall now show that the expression n(n + 1)(2n + 1) is an integral multiple of 6 ∀ n ∈ N. 

Assume T(n) = n(n+1)(2n+1) = 6x where x∈N. 

Basic Step: 

T(1) is true as shown above. 

Induction Step: 

Let T(k) be true for all k∈N. 

⇒ k (k +1) (2k +1) = 6x, where x∈N. .....(i) 

For T(k + 1), we replace k by (k + 1) in the given expression, i.e., 

T(k + 1) = (k + 1) (k + 2) (2 (k + 1) + 1) 

= (k + 1) (k + 2) ((2k + 1) + 2) 

= (k + 1) (k + 2) (2k + 1) + 2 (k + 1) (k + 2) 

= k (k + 1) (2k + 1) + 2 (k + 1) (2k + 1) + 2 (k + 1) (k + 2) 

= k (k + 1) (2k + 1) + 2 (k + 1) [(2k + 1) + (k + 2)] 

= 6x + 2(k + 1)(3k + 3) 

= 6x + 6(k + 1)2 

= 6 (x + (k + 1)2) = 6 × a positive integer 

∴ T(k) is true ⇒ T(k + 1) is true. 

∴ T(n) is true for all n∈N

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.

...