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In a certain sequence the term an is given by the formula: an=an−1+2n

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In a certain sequence the term an is given by the formula: an=an−1+2n  [#permalink]

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New post 15 Jun 2015, 10:39
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In a certain sequence the term an is given by the formula:

\(a_n=a_{n−1}+2n-1\) for all integer \(n>1\). If \(a_7=49\), what is \(a_1\)?

A. −1
B. 0
C. 1
D. 2
E. 3

The issue of this question is finding a rule for moving backwards from the 7th term to the 1st term. Isolate an-1:

an = an-1 + 2n - 1

--> an-1 = an − 2n + 1

Plug n into this relation to find the terms preceding the 7th term.

Plug a7=49, and n=7 into an-1 = an − 2n + 1 so:

a6=49−2×7+1=36 (Plugging in a7=49 and n=7)

a5=36−2×6+1=25 (Plugging in a6=36 and n=6)

Do you see the pattern? Each term an equals n2. Hence, a1 is 1.

If you missed the pattern, proceed to calculate a4, a3, a2, and a1:

a4=25−2×5+1=16 (Plugging in a5=25 and n=5)

a3=16−2×4+1=9 (Plugging in a4=16 and n=4)

a2=9−2×3+1=4 (Plugging in a3=9 and n=3)

a1=4−2×2+1=1 (Plugging in a2=4 and n=2)

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In a certain sequence the term an is given by the formula: an=an−1+2n  [#permalink]

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New post 15 Jun 2015, 11:03
[quote="reto"]In a certain sequence the term an is given by the formula:

\(a_n=a_{n−1}+2n-1\) for all integer \(n>1\). If \(a_7=49\), what is \(a_1\)?

A. −1
B. 0
C. 1
D. 2
E. 3

..
Hi,
\(a_n=a_{n−1}+2n-1\)....
as we can see that each subsequent term is 2n-1 greater than previous..
so we can straightway find \(a_7\)=(2*2-1)+(2*3-1)+(2*4-1)+(2*5-1)+(2*6-1)+(2*7-1)+\(a_1\)=3+5+7+9+11+13+\(a_1\)
\(a_7\)=49=48+\(a_1\)..
or \(a_1\)=1
ans C
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Re: In a certain sequence the term an is given by the formula: an=an−1+2n  [#permalink]

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New post 25 Jan 2019, 19:54
chetan2u wrote:
reto wrote:
In a certain sequence the term an is given by the formula:

\(a_n=a_{n−1}+2n-1\) for all integer \(n>1\). If \(a_7=49\), what is \(a_1\)?

A. −1
B. 0
C. 1
D. 2
E. 3

..
Hi,
\(a_n=a_{n−1}+2n-1\)....
as we can see that each subsequent term is 2n-1 greater than previous..
so we can straightway find \(a_7\)=(2*1-1)+(2*2-1)+(2*3-1)+(2*4-1)+(2*5-1)+(2*6-1)+(2*7-1)+\(a_1\)=3+5+7+9+11+13+\(a_1\)
\(a_7\)=49=48+\(a_1\)..
or \(a_1\)=1
ans C


Hello chetan2u !

Could you please explain to me which number corresponds to which of the letters?

\(a_n=a_{n−1}+2n-1\)

(2*1-1) I guess this is A2, but, where does the 2*1 is coming from?

Kind regards!
VP
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In a certain sequence the term an is given by the formula: an=an−1+2n  [#permalink]

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New post 25 Jan 2019, 20:33
reto wrote:
In a certain sequence the term an is given by the formula:

\(a_n=a_{n−1}+2n-1\) for all integer \(n>1\). If \(a_7=49\), what is \(a_1\)?
A. −1
B. 0
C. 1
D. 2
E. 3


If \(a_7=49\), just subustutute the value in the series and back solve it, while solving you will notice that for this part
a_n =a_{n−1}+2n-1 there will be a difference of 2
49 = a6 +13

36 = a5 + 11

25 = a4 + 9

16 = a3 + 7

9 = a2 + 5

4 = a1 + 3

a1 = 1

C
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Re: In a certain sequence the term an is given by the formula: an=an−1+2n  [#permalink]

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New post 25 Jan 2019, 23:19
1
1
jfranciscocuencag wrote:
chetan2u wrote:
reto wrote:
In a certain sequence the term an is given by the formula:

\(a_n=a_{n−1}+2n-1\) for all integer \(n>1\). If \(a_7=49\), what is \(a_1\)?

A. −1
B. 0
C. 1
D. 2
E. 3

..
Hi,
\(a_n=a_{n−1}+2n-1\)....
as we can see that each subsequent term is 2n-1 greater than previous..
so we can straightway find \(a_7\)=(2*1-1)+(2*2-1)+(2*3-1)+(2*4-1)+(2*5-1)+(2*6-1)+(2*7-1)+\(a_1\)=3+5+7+9+11+13+\(a_1\)
\(a_7\)=49=48+\(a_1\)..
or \(a_1\)=1
ans C


Hello chetan2u !

Could you please explain to me which number corresponds to which of the letters?

\(a_n=a_{n−1}+2n-1\)

(2*1-1) I guess this is A2, but, where does the 2*1 is coming from?

Kind regards!


Hi,
\(a_n=a_{n−1}+2n-1........a_2=a_1+2*2-1....a_3=a_2+2*3-1=a_1+2*2-1+2*3-1.......a_4=a_3+2*4-1=a_1+2*2-1+2*3-1+2*4-1\), therefore \(a_7=a_1+2*2-1+2*3-1+2*4-1+2*5-1+2*6-1+2*7-1\)...

But yes, 2*1-1 is typo and erroneously added, otherwise I have not added this in the later part 3+5+7+9+11+13+a1...
Thank you
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Re: In a certain sequence the term an is given by the formula: an=an−1+2n   [#permalink] 25 Jan 2019, 23:19
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