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

Question

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

Archit3110 · Answer

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

jfranciscocuencag · Answer

Hello  GMATinsight !

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

Regards!

GMATinsight · Answer

jfranciscocuencag

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!!!

bumpbot · Answer

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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

Prep Toolkit

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