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Bunuel
If n is the square of a positive integer, which of the following must be equal to the square of the next positive integer?

A. √n + 1
B. n + 1
C. n^2 + 1
D. n + 2√n + 1
E. n^2 + 2n + 1


next positive integer is sqrt(n)+1.
square of this number is [sqrt(n)+1]^2
the result is definitely a number with an "n" not squared not under radical
A,C, E out
B can't be the answer
so it must be D.

not even needed to square everything...
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Bunuel
If n is the square of a positive integer, which of the following must be equal to the square of the next positive integer?

A. √n + 1
B. n + 1
C. n^2 + 1
D. n + 2√n + 1
E. n^2 + 2n + 1

n is +ve integer

n = x^2 thus sqrt n = x , (x+1)^2 = x^2 + 2x+1 , substitute n^2+2sqrtn+1
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Bunuel
If n is the square of a positive integer, which of the following must be equal to the square of the next positive integer?

A. √n + 1
B. n + 1
C. n^2 + 1
D. n + 2√n + 1
E. n^2 + 2n + 1

Plug in some value and check -

Let n = 1 ( least number )
Square of the next positive integer is 4

Now, check the options : Only (D) and (E) satisfies

Check using one more square number to confirm the answer

Let n = 4 ( least number )
Square of the next positive integer is 9

Now, check the options : Only (D) satisfies

Hence answer must be option (D)
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Option D

\(n = K^2\); K>0 and Integer

\((K+1)^2 = K^2 + 2K + 1 = n + 2\sqrt{n} + 1\)
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Skywalker18
Let a= +ve integer such that
n= a^2
(a+1)= next +ve integer
(a+1)^2 = a^2 + 1 + 2a

and a=n^(1/2)
Therefore ,
(a+1)^2=n+1+2*n^(1/2)
Answer D

Alternatively,
we can take values
let n= 4 =a^2
=> a= 2
(a+1)=3^2 = 9
now checking the choices
A. √n + 1
= 2+1 =3
B. n + 1
= 4+1 =5
C. n^2 + 1
= 16+1 =17
D. n + 2√n + 1
= 4+ 2*2 +1 = 9
Answer D
E. n^2 + 2n + 1
= 16+ 8 + 1 =25


This question makes no sense to me . Isn't 5 the next positive integer after 4, and 5^2=25. I definitely did everything you prescribed on my first attempt at this question. Could you elaborate . Answer seems like it should be E to me . It always give you the square of the next positive integer . I am completely perplexed here. Think I'm just being dumb. Thanks !
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Skywalker18
Let a= +ve integer such that
n= a^2
(a+1)= next +ve integer
(a+1)^2 = a^2 + 1 + 2a

and a=n^(1/2)
Therefore ,
(a+1)^2=n+1+2*n^(1/2)
Answer D

Alternatively,
we can take values
let n= 4 =a^2
=> a= 2
(a+1)=3^2 = 9
now checking the choices
A. √n + 1
= 2+1 =3
B. n + 1
= 4+1 =5
C. n^2 + 1
= 16+1 =17
D. n + 2√n + 1
= 4+ 2*2 +1 = 9
Answer D
E. n^2 + 2n + 1
= 16+ 8 + 1 =25


This question makes no sense to me . Isn't 6 the next positive integer after 4, and 6^2=36. I definitely did everything you prescribed on my first attempt at this question. Could you elaborate . Think I'm just being dumb. Thanks !

If n is the square of a positive integer, which of the following must be equal to the square of the next positive integer?

A. \(\sqrt{n} + 1\)

B. \(n + 1\)

C. \(n^2 + 1\)

D. \(n + 2\sqrt{n} + 1\)

E. \(n^2 + 2n + 1\)

n is the square of a positive integer. Say that positive integer is x. So, we are given that \(n = x^2\). Notice that from this it follows that \(x=\sqrt{n}\)

Which of the following must be equal to the square of the next positive integer?

The next positive integer is x + 1. Its square = \((x+1)^2=x^2+2x+1\).

Substitute the values of x^2 and x from above to get: \((x+1)^2=x^2+2x+1=n+2\sqrt{n}+1\).

Answer: D.
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Bunuel
If n is the square of a positive integer, which of the following must be equal to the square of the next positive integer?

A. \(\sqrt{n} + 1\)

B. \(n + 1\)

C. \(n^2 + 1\)

D. \(n + 2\sqrt{n} + 1\)

E. \(n^2 + 2n + 1\)

We can let n = k^2 where k is a positive integer, we see that the square of the next positive integer is
(k + 1)^2 = k^2 + 2k + 1. Since n = k^2, so k = √n. Therefore, in terms of n, the square of the next positive integer is n + 2√n + 1.

Answer: D
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