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Math Expert V
Joined: 02 Sep 2009
Posts: 59725
Let ak be a sequence of integers such that a1 = 1 and a(m+n) =am + an  [#permalink]

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Difficulty:   35% (medium)

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

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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

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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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1
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
Senior Manager  G
Joined: 12 Sep 2017
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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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Hello GMATinsight!

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

Regards!
CEO  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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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. $$a_{12} = 1+2+3+4+5+6+7+8+9+10+11+12 = (\frac{1}{2})(12)*(12+1) = 6*13 = 78$$

I hope this helps!!!
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