Mixing
Arithmetic series problem
Clash Royale CLAN TAG#URR8PPP
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Given $lefta_nright$ arithmetic progression, $a_1=2$, $a_n+1=a_n+2n$ $left(n:ge :1right)$. $a_50=?$
What i did:
$$a_n+d=a_n+2n$$
$$d=2n$$
$$a_50=2+dleft(n-1right)$$
$$a_50=2+2left(n^2-nright)$$
$$a_50=2+2cdot 2450$$
$$a_50=4902$$
But this is wrong. Answers:
$$A=2452,:B=2450,:C=2552,:D=2500$$
sequences-and-series
asked Jul 31 at 15:06
Serif Yaohim
233
add a comment |Â
up vote
4
down vote
favorite
Given $lefta_nright$ arithmetic progression, $a_1=2$, $a_n+1=a_n+2n$ $left(n:ge :1right)$. $a_50=?$
What i did:
$$a_n+d=a_n+2n$$
$$d=2n$$
$$a_50=2+dleft(n-1right)$$
$$a_50=2+2left(n^2-nright)$$
$$a_50=2+2cdot 2450$$
$$a_50=4902$$
But this is wrong. Answers:
$$A=2452,:B=2450,:C=2552,:D=2500$$
sequences-and-series
asked Jul 31 at 15:06
Serif Yaohim
233
1
This is not an arithmetic progression.
– Yves Daoust
Jul 31 at 15:20
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given $lefta_nright$ arithmetic progression, $a_1=2$, $a_n+1=a_n+2n$ $left(n:ge :1right)$. $a_50=?$
What i did:
$$a_n+d=a_n+2n$$
$$d=2n$$
$$a_50=2+dleft(n-1right)$$
$$a_50=2+2left(n^2-nright)$$
$$a_50=2+2cdot 2450$$
$$a_50=4902$$
But this is wrong. Answers:
$$A=2452,:B=2450,:C=2552,:D=2500$$
sequences-and-series
asked Jul 31 at 15:06
Serif Yaohim
233
Given $lefta_nright$ arithmetic progression, $a_1=2$, $a_n+1=a_n+2n$ $left(n:ge :1right)$. $a_50=?$
What i did:
$$a_n+d=a_n+2n$$
$$d=2n$$
$$a_50=2+dleft(n-1right)$$
$$a_50=2+2left(n^2-nright)$$
$$a_50=2+2cdot 2450$$
$$a_50=4902$$
But this is wrong. Answers:
$$A=2452,:B=2450,:C=2552,:D=2500$$
sequences-and-series
asked Jul 31 at 15:06
Serif Yaohim
233
asked Jul 31 at 15:06
Serif Yaohim
233
asked Jul 31 at 15:06
Serif Yaohim
233
asked Jul 31 at 15:06
Serif Yaohim
233
233
1
This is not an arithmetic progression.
– Yves Daoust
Jul 31 at 15:20
add a comment |Â
1
This is not an arithmetic progression.
– Yves Daoust
Jul 31 at 15:20
1
1
This is not an arithmetic progression.
– Yves Daoust
Jul 31 at 15:20
This is not an arithmetic progression.
– Yves Daoust
Jul 31 at 15:20
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Summing up the expressions $a_n+1 = a_n + 2n$ for $n = 1, dots, k$ one easily gets that
$$
a_k+1 = a_1 + sum_i=1^k 2i = 2 + k(k+1)
$$
Hence, $a_50 = 2 + 2450 = 2452$ and the correct answer is A.
answered Jul 31 at 15:13
pointguard0
689517
add a comment |Â
up vote
5
down vote
beginaligna_1&=2\a_n+1&=a_n+2n\&=a_n-1+2[n+(n-1)]\&=a_n-2+2[n+(n-1)+(n-2)]\
&= ldots\
&=a_1+2[n+(n-1)+ldots +1]endalign
Hence beginaligna_50&=a_1+2(49+ldots + 1)\&=2+2cdot frac49(50)2\&=2+49(50)\&=2+(50-1)(50)\&=2+2500-50\&=2452 endalign
Remark:
This is not an AP, if it is an AP, the difference between consecutive terms is a constant.
answered Jul 31 at 15:11
Siong Thye Goh
76.7k134794
I didn't understand second part please help
– Serif Yaohim
Jul 31 at 17:03
I just let n=49
– Siong Thye Goh
Aug 1 at 3:04
add a comment |Â
up vote
2
down vote
Alternatively, note: $a_n+1-a_n=2n$. So:
$$beginalign a_2-a_1&=2cdot 1\
a_3-a_2&=2cdot 2\
a_4-a_3&=2cdot 3\
&vdots \
a_50-a_49&=2cdot 49 endalign$$
Summing all (midterms telescope):
$$a_50-a_1=2(1+2+3+cdots +49) Rightarrow \
a_50=a_1+2cdot frac1+492cdot 49=2+2450=2452.$$
answered Jul 31 at 18:53
farruhota
13.5k2632
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Summing up the expressions $a_n+1 = a_n + 2n$ for $n = 1, dots, k$ one easily gets that
$$
a_k+1 = a_1 + sum_i=1^k 2i = 2 + k(k+1)
$$
Hence, $a_50 = 2 + 2450 = 2452$ and the correct answer is A.
answered Jul 31 at 15:13
pointguard0
689517
add a comment |Â
up vote
3
down vote
accepted
Summing up the expressions $a_n+1 = a_n + 2n$ for $n = 1, dots, k$ one easily gets that
$$
a_k+1 = a_1 + sum_i=1^k 2i = 2 + k(k+1)
$$
Hence, $a_50 = 2 + 2450 = 2452$ and the correct answer is A.
answered Jul 31 at 15:13
pointguard0
689517
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Summing up the expressions $a_n+1 = a_n + 2n$ for $n = 1, dots, k$ one easily gets that
$$
a_k+1 = a_1 + sum_i=1^k 2i = 2 + k(k+1)
$$
Hence, $a_50 = 2 + 2450 = 2452$ and the correct answer is A.
answered Jul 31 at 15:13
pointguard0
689517
Summing up the expressions $a_n+1 = a_n + 2n$ for $n = 1, dots, k$ one easily gets that
$$
a_k+1 = a_1 + sum_i=1^k 2i = 2 + k(k+1)
$$
Hence, $a_50 = 2 + 2450 = 2452$ and the correct answer is A.
answered Jul 31 at 15:13
pointguard0
689517
answered Jul 31 at 15:13
pointguard0
689517
answered Jul 31 at 15:13
pointguard0
689517
answered Jul 31 at 15:13
pointguard0
689517
689517
add a comment |Â
add a comment |Â
up vote
5
down vote
beginaligna_1&=2\a_n+1&=a_n+2n\&=a_n-1+2[n+(n-1)]\&=a_n-2+2[n+(n-1)+(n-2)]\
&= ldots\
&=a_1+2[n+(n-1)+ldots +1]endalign
Hence beginaligna_50&=a_1+2(49+ldots + 1)\&=2+2cdot frac49(50)2\&=2+49(50)\&=2+(50-1)(50)\&=2+2500-50\&=2452 endalign
Remark:
This is not an AP, if it is an AP, the difference between consecutive terms is a constant.
answered Jul 31 at 15:11
Siong Thye Goh
76.7k134794
I didn't understand second part please help
– Serif Yaohim
Jul 31 at 17:03
I just let n=49
– Siong Thye Goh
Aug 1 at 3:04
add a comment |Â
up vote
5
down vote
beginaligna_1&=2\a_n+1&=a_n+2n\&=a_n-1+2[n+(n-1)]\&=a_n-2+2[n+(n-1)+(n-2)]\
&= ldots\
&=a_1+2[n+(n-1)+ldots +1]endalign
Hence beginaligna_50&=a_1+2(49+ldots + 1)\&=2+2cdot frac49(50)2\&=2+49(50)\&=2+(50-1)(50)\&=2+2500-50\&=2452 endalign
Remark:
This is not an AP, if it is an AP, the difference between consecutive terms is a constant.
answered Jul 31 at 15:11
Siong Thye Goh
76.7k134794
I didn't understand second part please help
– Serif Yaohim
Jul 31 at 17:03
I just let n=49
– Siong Thye Goh
Aug 1 at 3:04
add a comment |Â
up vote
5
down vote
up vote
5
down vote
beginaligna_1&=2\a_n+1&=a_n+2n\&=a_n-1+2[n+(n-1)]\&=a_n-2+2[n+(n-1)+(n-2)]\
&= ldots\
&=a_1+2[n+(n-1)+ldots +1]endalign
Hence beginaligna_50&=a_1+2(49+ldots + 1)\&=2+2cdot frac49(50)2\&=2+49(50)\&=2+(50-1)(50)\&=2+2500-50\&=2452 endalign
Remark:
This is not an AP, if it is an AP, the difference between consecutive terms is a constant.
answered Jul 31 at 15:11
Siong Thye Goh
76.7k134794
beginaligna_1&=2\a_n+1&=a_n+2n\&=a_n-1+2[n+(n-1)]\&=a_n-2+2[n+(n-1)+(n-2)]\
&= ldots\
&=a_1+2[n+(n-1)+ldots +1]endalign
Hence beginaligna_50&=a_1+2(49+ldots + 1)\&=2+2cdot frac49(50)2\&=2+49(50)\&=2+(50-1)(50)\&=2+2500-50\&=2452 endalign
Remark:
This is not an AP, if it is an AP, the difference between consecutive terms is a constant.
answered Jul 31 at 15:11
Siong Thye Goh
76.7k134794
edited Jul 31 at 15:20
answered Jul 31 at 15:11
Siong Thye Goh
76.7k134794
answered Jul 31 at 15:11
Siong Thye Goh
76.7k134794
answered Jul 31 at 15:11
Siong Thye Goh
76.7k134794
76.7k134794
I didn't understand second part please help
– Serif Yaohim
Jul 31 at 17:03
I just let n=49
– Siong Thye Goh
Aug 1 at 3:04
add a comment |Â
I didn't understand second part please help
– Serif Yaohim
Jul 31 at 17:03
I just let n=49
– Siong Thye Goh
Aug 1 at 3:04
I didn't understand second part please help
– Serif Yaohim
Jul 31 at 17:03
I didn't understand second part please help
– Serif Yaohim
Jul 31 at 17:03
I just let n=49
– Siong Thye Goh
Aug 1 at 3:04
I just let n=49
– Siong Thye Goh
Aug 1 at 3:04
add a comment |Â
up vote
2
down vote
Alternatively, note: $a_n+1-a_n=2n$. So:
$$beginalign a_2-a_1&=2cdot 1\
a_3-a_2&=2cdot 2\
a_4-a_3&=2cdot 3\
&vdots \
a_50-a_49&=2cdot 49 endalign$$
Summing all (midterms telescope):
$$a_50-a_1=2(1+2+3+cdots +49) Rightarrow \
a_50=a_1+2cdot frac1+492cdot 49=2+2450=2452.$$
answered Jul 31 at 18:53
farruhota
13.5k2632
add a comment |Â
up vote
2
down vote
Alternatively, note: $a_n+1-a_n=2n$. So:
$$beginalign a_2-a_1&=2cdot 1\
a_3-a_2&=2cdot 2\
a_4-a_3&=2cdot 3\
&vdots \
a_50-a_49&=2cdot 49 endalign$$
Summing all (midterms telescope):
$$a_50-a_1=2(1+2+3+cdots +49) Rightarrow \
a_50=a_1+2cdot frac1+492cdot 49=2+2450=2452.$$
answered Jul 31 at 18:53
farruhota
13.5k2632
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Alternatively, note: $a_n+1-a_n=2n$. So:
$$beginalign a_2-a_1&=2cdot 1\
a_3-a_2&=2cdot 2\
a_4-a_3&=2cdot 3\
&vdots \
a_50-a_49&=2cdot 49 endalign$$
Summing all (midterms telescope):
$$a_50-a_1=2(1+2+3+cdots +49) Rightarrow \
a_50=a_1+2cdot frac1+492cdot 49=2+2450=2452.$$
answered Jul 31 at 18:53
farruhota
13.5k2632
Alternatively, note: $a_n+1-a_n=2n$. So:
$$beginalign a_2-a_1&=2cdot 1\
a_3-a_2&=2cdot 2\
a_4-a_3&=2cdot 3\
&vdots \
a_50-a_49&=2cdot 49 endalign$$
Summing all (midterms telescope):
$$a_50-a_1=2(1+2+3+cdots +49) Rightarrow \
a_50=a_1+2cdot frac1+492cdot 49=2+2450=2452.$$
answered Jul 31 at 18:53
farruhota
13.5k2632
answered Jul 31 at 18:53
farruhota
13.5k2632
answered Jul 31 at 18:53
farruhota
13.5k2632
answered Jul 31 at 18:53
farruhota
13.5k2632
13.5k2632
add a comment |Â
add a comment |Â
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1
This is not an arithmetic progression.
– Yves Daoust
Jul 31 at 15:20