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Let ak be a sequence of integers such that a1 = 1 and a(m+n) =am + an

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Let ak be a sequence of integers such that a1 = 1 and a(m+n) =am + an  [#permalink]

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New post 19 Mar 2019, 01:33
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

71% (02:34) correct 29% (03:06) wrong based on 31 sessions

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Re: Let ak be a sequence of integers such that a1 = 1 and a(m+n) =am + an  [#permalink]

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New post 19 Mar 2019, 03:44
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Bunuel wrote:
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


a1=1

use relation
\(a_{m+n}=a_m+a_n+mn\)
a1+1 = 1+1+1
a2=3
similarly
a2+1 = 1+3+2 ; 6= a3
a4=10, a5=15; a6= 21
so
a12 will be a6+6 = 21+21+36 = 78
IMO D
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Let ak be a sequence of integers such that a1 = 1 and a(m+n) =am + an  [#permalink]

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New post 13 Apr 2019, 21:54
Hello GMATinsight!

Could you please help me also with this one? Ive been trying almost for an hour =/

Regards!
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Let ak be a sequence of integers such that a1 = 1 and a(m+n) =am + an  [#permalink]

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New post 13 Apr 2019, 22:34
1
jfranciscocuencag wrote:
Hello GMATinsight!

Could you please help me also with this one? Ive been trying almost for an hour =/

Regards!


jfranciscocuencag

Quote:
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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Let ak be a sequence of integers such that a1 = 1 and a(m+n) =am + an   [#permalink] 13 Apr 2019, 22:34
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