jfranciscocuencag wrote:
Hello
GMATinsight!
Could you please help me also with this one? Ive been trying almost for an hour =/
Regards!
jfranciscocuencagQuote:
Let \(a_k\) be a sequence of integers such that \(a_1=1\) and \(a_{m+n}=a_m+a_n+mn\), for all positive integers m and n. Then \(a_{12}\) is
(A) 45
(B) 56
(C) 67
(D) 78
(E) 89
The simplest way to solve questions of series is to write a few terms of the series and analyse the pattern among those terms
Given: \(a_1=1\) and \(a_{m+n}=a_m+a_n+mn\)
i.e. \(a_2 = a_{1+1} = a_1 + a_1 + 1*1 = 1+1+1 = 3\)
Similarly, \(a_3 = a_{1+2} = a_1 + a_2 + 1*2 = 1+3+2 = 6\)
Similarly, \(a_4 = a_{1+3} = a_1 + a_3 + 1*3 = 1+6+3 = 10\)
Now, we see that series is progressing like {1, 3, 6, 10, ....}
The most important part here is to be able to visualise the pattern which is
\(a_1 = 1\)
\(a_2 = 1+2\)
\(a_3 = 1+2+3\)
\(a_4 = 1+2+3+4\)... and so on...
.....
.....
i.e. \(a_{12} = 1+2+3+4+5+6+7+8+9+10+11+12\)
CONCEPT: Sum of first n distinct positive integers i.e. \(1+2+3+4.....+n = (\frac{1}{2})*n*(n+1)\)i.e. i.e. \(a_{12} = 1+2+3+4+5+6+7+8+9+10+11+12 = (\frac{1}{2})(12)*(12+1) = 6*13 = 78\)
Answer: Option D
I hope this helps!!!
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