In the sequence a1,a2,…,an,…, an=n2 for all n>0. For what positive int

Hi All,

chetan2u found the "secret" to this question - while the prompt is based in Algebra, the 'shortcut' is actually based in Geometry (specifically a multiple of the 3/4/5 right triangle). You do not need to find the shortcut to answer this question though, if you know your Perfect Squares and how to do basic arithmetic.

The notation in the prompt tells us that each term in the sequence is equal to the "number" of the term squared.

So....
A1 = 1^2 = 1
A2 = 2^2 = 4
A3 = 3^2 = 9
Etc.

We're asked to find the value of N that completes the given equation. Since the answers ARE numbers, we can TEST THE ANSWERS until we find the match.

IF N=7
Does the 5th term + 8th term = 11th term????
25 + 64 is NOT 121

IF N = 8
Does the 6th term + 9th term = 12th term????
36 + 81 is NOT 144

IF N = 9
Does the 7th term + 10th term = 13th term???
49 + 100 is NOT 169

IF N = 10
Does the 8th term + 11th term = 14th term????
64 + 121 is NOT 196

We've eliminated 4 answers, so the remaining answer MUST be correct. Here's the proof though:

IF N = 11
Does the 9th term + 12th term = 15th term???

81 + 144 DOES = 225

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Answer = E = 11

\((n-2)^2 + (n-1)^2 = (n+4)^2\)

\((n-2+n+4)(n-2-n-4) + (n-1)^2 = 0\)

\(-(2n-2)6 + (n-1)^2 = 0\)

n = 11

Just expanding both sides we get (n-2)^2 + (n+1)^2 = (n+4)^2
Expanding and cancelling terms we get n^2-10n-11 = 0
Or (n-11)(n+1) = 0
So n=11 or -1.
Since n is positive, n=11.

Answer = E.

Press kudos if correct.

(n-2)^2 +(n-1)^2=(n-4)^2
(n-4)^2-(n-2)^2=(n-1)^2
(n-4-n+2)(n-4+n-2)=(n-1)^2
-2(2n-6)=(n-1)^2
n=11

VERITAS PREP OFFICIAL SOLUTION:

One method to this problem is starting with n=7 and just plowing through the arithmetic of the answer choices. (Does 5^2 + 8^2 = 11^2? Nope. Does 6^2 + 9^2 = 12^2? Nope. And so forth.) This can get a bit tedious, though.

Algebra offers a better way. Since \(a_{n−2}=(n–2)^2\), \(a_{n+1}=(n+1)^2\), and \(a_{n+4}=(n+4)^2\), the statement \(a_{n−2}+a_{n+1}=a_{n+4}\) can be rewritten as \((n–2)^2+(n+1)^2=(n+4)^2\). Expanding these common algebraic equations gives

\((n^2–4n+4)+(n^2+2n+1)=n^2+8n+16\)

Simplifying gives

\(2n^2−2n+5=n^2+8n+16\)

Moving everything to the left side produces the quadratic equation

\(n^2–10n–11=0\)

Factoring this equation produces

\((n–11)(n+1)=0\)

So n = 11 or –1. We know that n is a positive integer, so it must be that n = 11.

Better still, enterprising students may note the 9, 12, 15. Pythagorean triple lurking in this problem. Identifying the a^2 + b^2 = c^2 structure of this problem and searching for a triple that fits can drastically shortcut the work required here.

30 second approach:

Since the question is asking for the value of n and because the answer choices are small enough to work with, simply plug them in for n to check whether the equation holds.