The square of any odd number is of the form - (a) 8n+1 (b) 6n+1 (c) 4n+1 (d) 2n+1. The answer is a,c & d

Question

how do i get options - a,c and d as the answer?
plz solve this problem. The question is in the chapter BINOMIAL THEOREM, so prefferably do it by a method related to that.
BUT u may use any method.
thanks a lot for helping me.

Answer 1

Let m = any integer
2m + 1 = odd number
square of odd number = (2m + 1)^2 = 4m^2 + 4m + 1 = 4(m^2 + m) + 1

Case 1: m^2 + m = n, then (c) is answer
Case 2: 2(m^2 + m) = n, then (d) is answer

Case 3:
Next, m^2 + m = even

Why?
Because if m is odd then m^2 is also odd so (m^2 + m) is sum of two odd numbers = even number
But if m is even then m^2 is also even so (m^2 + m) is sum of two even numbers = even number

As (m^2 + m) is even so let (m^2 + m) = 2n [n is an integer].
Thus (2m+1)^2 = 8n + 1, so (a) is the answer

Answer 2

Let x be an odd number. Then x is of the form

x = 2k - 1. It follows that

x^2 = (2k - 1)^2 = 4k^2 - 4k + 1

(a) To get this in the form 8n + 1, note that

4k^2 - 4k + 1 = 4(k)(k - 1) + 1

Since k(k - 1) is an odd number times an even umber, it follows that k(k - 1) is even, so k(k - 1) = 2n, and
4(2n) + 1 = 8n + 1.

(b) It cannot be put in this form.

(c) As calculated earlier, (4k - 1)^2 = 4k^2 - 4k + 1, so if we factor 4 out of the first two terms,

4(k^2 - k) + 1, which is in the form 4n + 1 (if we let n = k^2 - k).

d) 4k^2 - 4k + 1, so

2(2k^2 - 2k) + 1, and if we let n = 2k^2 - 2k, then our form is

2n + 1

Answer 3

Any odd number can be written as 2p + 1
so squaring gives
4p^2+4p + 1
=4p(p+1) + 1
Now either p or p+1 must be even, and so 4p(p+1) is equal to 8n for some integer n = p(p+1)/2. That gives (a)

Or put n=p(p+1), and it's (c) 4n+1

Or put n = 2p(p+1), and it's (d) 2n+1

I'm not answering any more: Puggy always gets the same answer ahead of me!!!!!

Answer 4

let n, m, x be integers
Let x be any odd number.

then x = 2n+1 { for some integer "n"}

x^2 = (2n + 1)^2 = 4n^2 + 4n + 1 ... here make note of the fact
that (a) (b) (c) and (d) have the { + 1} fragment at the end .

soooo ...

focus on the "4n^2 + 4n" portion....

a)
4n^2 + 4n = 8m {substituted "m" for "n"}
4n(n+1) = 8m
... if "n" is even, then 4n is divisible by 8
... if "n" is odd, then(n+1) is even (divisible by 2)
and "4n" is divisible by 4
so the product is divisible by 8
qed { ie "a" works }

c) 4n(n+1) = 4m {factor out a "4"}
... m = n(n+1)
qed { ie "c" works }

d) 4n(n+1) = 2m
... m = 2n(n +1)
qed { ie "d" works }

.........
however
b) 4n(n + 1) = 6m ... factor out a "2"
2n (n + 1) = 3m
then it MUST be the case that "2n" or "(n + 1)"
is divisible by "3"
let n = 1
2n = 2 {and NOT divisible by 3}
(n + 1) = 2 {and NOT divisible by 3}

so, by counterexample "b" does not work.


=> (a) (c) and (d) are "answers" and (b) is not.

Answer 5

Just one comment about Puggy's solution:

(c) and (d) are more easily found by reindexing 8n+1 (yes, you should be allowed to do that), since 4(2n)+1 and 2(4n)+1 are the correct forms for (c) and (d) respectively.

Answer 6

Let's use the number 9. Square of the odd number 3.

8N + 1 = 9
8N = 8
N = 1
Divides evenly.

4N + 1 = 9
4N = 8
N = 2
Divides evenly

2N + 1 = 9
2N = 8
N = 4
Divides evenly.

HOWEVER, for B, this is not the case:
6N + 1 = 9
6N = 8
N = 4/3

A, B, and C will alway divide evenly, so long as the number is the square of an odd number. B only sometimes will work, for example, with 25.

Answer 7

suppose odd number m=2k+1(k:N)
then, m^2=(2k+1)^2=
=4k^2+4k+1=
=4k(k+1)+1
from the above formula , answer c and d are clear

next, (a)
k(k+1) is always even number
because one of k and (k+1) is always even number.
that is, if k is even number , then , of course, k(k+1) is even number and if k is odd number, then k+1 is even number
so 4k(k+1)=8n(n:N)
hence answer (a) is just right!