Compute the value of the sum $7cdot 11+11cdot 15+15cdot 19+cdots+95cdot 99$
$begingroup$
Solution: My attempt:
=$9^2-2^2 +13^2-2^2+17^2-2^2...97^2-2^2$
=$(9^2+13^2+17^2...97^2)-2^2( 23 )$
Focus on the first part. 81+169+289+441...Here first-order differences are 88, 120, 152... and the second-order difference is common to be 32. Therefore the series is of the form of $p(n)^2+q(n)+r$. If we plug n=0, we get the first term, if plug n=1, we get the second term, if we plug n=2, we get the third term and so on.
$p*0^2+q*0+r=81$
r=81
$p*1^2*q*1+81=169$
$p+q+81=169$
$p+q=88…(1)$
$p*2^2+q*2+81=289$
$4p+2q=208…(2)$
Solving (1) and (2), we get p =16, q =72, and r =81.
Therefore the general term formula is $16n^2+72n+81$. Now, if we start adding them up, 81 will be there 23 times. Then there will be 72, 144, 216 etc, 22 times. This sum will be 72(1+2+3+...22) =18216. At the last 16n^2 will be there 22 times. This sum will be $16(1^2+2^2+3^2...22^2) = 16 ((22) (23) (45)/6) = 60720$. Grand total is $1863+18216+60720-92= 80707$.
arithmetic-progressions
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
asked Dec 19 '18 at 14:26
Sherlock WatsonSherlock Watson
3462413
$endgroup$
add a comment |
$begingroup$
Solution: My attempt:
=$9^2-2^2 +13^2-2^2+17^2-2^2...97^2-2^2$
=$(9^2+13^2+17^2...97^2)-2^2( 23 )$
Focus on the first part. 81+169+289+441...Here first-order differences are 88, 120, 152... and the second-order difference is common to be 32. Therefore the series is of the form of $p(n)^2+q(n)+r$. If we plug n=0, we get the first term, if plug n=1, we get the second term, if we plug n=2, we get the third term and so on.
$p*0^2+q*0+r=81$
r=81
$p*1^2*q*1+81=169$
$p+q+81=169$
$p+q=88…(1)$
$p*2^2+q*2+81=289$
$4p+2q=208…(2)$
Solving (1) and (2), we get p =16, q =72, and r =81.
Therefore the general term formula is $16n^2+72n+81$. Now, if we start adding them up, 81 will be there 23 times. Then there will be 72, 144, 216 etc, 22 times. This sum will be 72(1+2+3+...22) =18216. At the last 16n^2 will be there 22 times. This sum will be $16(1^2+2^2+3^2...22^2) = 16 ((22) (23) (45)/6) = 60720$. Grand total is $1863+18216+60720-92= 80707$.
arithmetic-progressions
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
asked Dec 19 '18 at 14:26
Sherlock WatsonSherlock Watson
3462413
$endgroup$
$begingroup$
$15cdot 19=17^2-2^2$
$endgroup$
– giannispapav
Dec 19 '18 at 14:34
$begingroup$
The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
$endgroup$
– user170231
Dec 19 '18 at 14:43
add a comment |
$begingroup$
Solution: My attempt:
=$9^2-2^2 +13^2-2^2+17^2-2^2...97^2-2^2$
=$(9^2+13^2+17^2...97^2)-2^2( 23 )$
Focus on the first part. 81+169+289+441...Here first-order differences are 88, 120, 152... and the second-order difference is common to be 32. Therefore the series is of the form of $p(n)^2+q(n)+r$. If we plug n=0, we get the first term, if plug n=1, we get the second term, if we plug n=2, we get the third term and so on.
$p*0^2+q*0+r=81$
r=81
$p*1^2*q*1+81=169$
$p+q+81=169$
$p+q=88…(1)$
$p*2^2+q*2+81=289$
$4p+2q=208…(2)$
Solving (1) and (2), we get p =16, q =72, and r =81.
Therefore the general term formula is $16n^2+72n+81$. Now, if we start adding them up, 81 will be there 23 times. Then there will be 72, 144, 216 etc, 22 times. This sum will be 72(1+2+3+...22) =18216. At the last 16n^2 will be there 22 times. This sum will be $16(1^2+2^2+3^2...22^2) = 16 ((22) (23) (45)/6) = 60720$. Grand total is $1863+18216+60720-92= 80707$.
arithmetic-progressions
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
asked Dec 19 '18 at 14:26
Sherlock WatsonSherlock Watson
3462413
$endgroup$
Solution: My attempt:
=$9^2-2^2 +13^2-2^2+17^2-2^2...97^2-2^2$
=$(9^2+13^2+17^2...97^2)-2^2( 23 )$
Focus on the first part. 81+169+289+441...Here first-order differences are 88, 120, 152... and the second-order difference is common to be 32. Therefore the series is of the form of $p(n)^2+q(n)+r$. If we plug n=0, we get the first term, if plug n=1, we get the second term, if we plug n=2, we get the third term and so on.
$p*0^2+q*0+r=81$
r=81
$p*1^2*q*1+81=169$
$p+q+81=169$
$p+q=88…(1)$
$p*2^2+q*2+81=289$
$4p+2q=208…(2)$
Solving (1) and (2), we get p =16, q =72, and r =81.
Therefore the general term formula is $16n^2+72n+81$. Now, if we start adding them up, 81 will be there 23 times. Then there will be 72, 144, 216 etc, 22 times. This sum will be 72(1+2+3+...22) =18216. At the last 16n^2 will be there 22 times. This sum will be $16(1^2+2^2+3^2...22^2) = 16 ((22) (23) (45)/6) = 60720$. Grand total is $1863+18216+60720-92= 80707$.
arithmetic-progressions
arithmetic-progressions
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
asked Dec 19 '18 at 14:26
Sherlock WatsonSherlock Watson
3462413
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
asked Dec 19 '18 at 14:26
Sherlock WatsonSherlock Watson
3462413
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
1,8602520
asked Dec 19 '18 at 14:26
Sherlock WatsonSherlock Watson
3462413
asked Dec 19 '18 at 14:26
Sherlock WatsonSherlock Watson
3462413
asked Dec 19 '18 at 14:26
Sherlock WatsonSherlock Watson
3462413
3462413
$begingroup$
$15cdot 19=17^2-2^2$
$endgroup$
– giannispapav
Dec 19 '18 at 14:34
$begingroup$
The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
$endgroup$
– user170231
Dec 19 '18 at 14:43
add a comment |
$begingroup$
$15cdot 19=17^2-2^2$
$endgroup$
– giannispapav
Dec 19 '18 at 14:34
$begingroup$
The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
$endgroup$
– user170231
Dec 19 '18 at 14:43
$begingroup$
$15cdot 19=17^2-2^2$
$endgroup$
– giannispapav
Dec 19 '18 at 14:34
$begingroup$
$15cdot 19=17^2-2^2$
$endgroup$
– giannispapav
Dec 19 '18 at 14:34
$begingroup$
The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
$endgroup$
– user170231
Dec 19 '18 at 14:43
$begingroup$
The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
$endgroup$
– user170231
Dec 19 '18 at 14:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have one more way to do it easily. As we can see the sequence has the general term $T_n$=$(4n+3)(4n+7)$ and the sum of sequence will be equal to $sum {T_n} $ from $n=1$ to $n=23$. Now, the sum will appear as :
$$16frac{n(n+1)(2n+1)}{6}+20n(n+1)+21n$$ where putting n=23 will give $80707$
It will be easier as there will be less chance for error as there will be less steps than what you have approached the question. Also what fellow users have given, it is just the same but easier coefficients to apply for.
answered Dec 19 '18 at 14:45
jayant98jayant98
653318
$endgroup$
$begingroup$
Thanks. How do you reach the general term? With induction?
$endgroup$
– Sherlock Watson
Dec 19 '18 at 18:28
$begingroup$
I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
$endgroup$
– jayant98
Dec 19 '18 at 18:40
add a comment |
$begingroup$
begin{align}
S &= sum_{n=0}^{n=22} (4n+7)(4n+11)\
&= sum_{n=0}^{n=22} 16n^2+72n+color{red}{77}\
&= 80707
end{align}
answered Dec 19 '18 at 14:40
Karn WatcharasupatKarn Watcharasupat
3,9742526
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have one more way to do it easily. As we can see the sequence has the general term $T_n$=$(4n+3)(4n+7)$ and the sum of sequence will be equal to $sum {T_n} $ from $n=1$ to $n=23$. Now, the sum will appear as :
$$16frac{n(n+1)(2n+1)}{6}+20n(n+1)+21n$$ where putting n=23 will give $80707$
It will be easier as there will be less chance for error as there will be less steps than what you have approached the question. Also what fellow users have given, it is just the same but easier coefficients to apply for.
answered Dec 19 '18 at 14:45
jayant98jayant98
653318
$endgroup$
$begingroup$
Thanks. How do you reach the general term? With induction?
$endgroup$
– Sherlock Watson
Dec 19 '18 at 18:28
$begingroup$
I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
$endgroup$
– jayant98
Dec 19 '18 at 18:40
add a comment |
$begingroup$
We have one more way to do it easily. As we can see the sequence has the general term $T_n$=$(4n+3)(4n+7)$ and the sum of sequence will be equal to $sum {T_n} $ from $n=1$ to $n=23$. Now, the sum will appear as :
$$16frac{n(n+1)(2n+1)}{6}+20n(n+1)+21n$$ where putting n=23 will give $80707$
It will be easier as there will be less chance for error as there will be less steps than what you have approached the question. Also what fellow users have given, it is just the same but easier coefficients to apply for.
answered Dec 19 '18 at 14:45
jayant98jayant98
653318
$endgroup$
$begingroup$
Thanks. How do you reach the general term? With induction?
$endgroup$
– Sherlock Watson
Dec 19 '18 at 18:28
$begingroup$
I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
$endgroup$
– jayant98
Dec 19 '18 at 18:40
add a comment |
$begingroup$
We have one more way to do it easily. As we can see the sequence has the general term $T_n$=$(4n+3)(4n+7)$ and the sum of sequence will be equal to $sum {T_n} $ from $n=1$ to $n=23$. Now, the sum will appear as :
$$16frac{n(n+1)(2n+1)}{6}+20n(n+1)+21n$$ where putting n=23 will give $80707$
It will be easier as there will be less chance for error as there will be less steps than what you have approached the question. Also what fellow users have given, it is just the same but easier coefficients to apply for.
answered Dec 19 '18 at 14:45
jayant98jayant98
653318
$endgroup$
We have one more way to do it easily. As we can see the sequence has the general term $T_n$=$(4n+3)(4n+7)$ and the sum of sequence will be equal to $sum {T_n} $ from $n=1$ to $n=23$. Now, the sum will appear as :
$$16frac{n(n+1)(2n+1)}{6}+20n(n+1)+21n$$ where putting n=23 will give $80707$
It will be easier as there will be less chance for error as there will be less steps than what you have approached the question. Also what fellow users have given, it is just the same but easier coefficients to apply for.
answered Dec 19 '18 at 14:45
jayant98jayant98
653318
edited Dec 19 '18 at 14:51
answered Dec 19 '18 at 14:45
jayant98jayant98
653318
answered Dec 19 '18 at 14:45
jayant98jayant98
653318
answered Dec 19 '18 at 14:45
jayant98jayant98
653318
653318
$begingroup$
Thanks. How do you reach the general term? With induction?
$endgroup$
– Sherlock Watson
Dec 19 '18 at 18:28
$begingroup$
I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
$endgroup$
– jayant98
Dec 19 '18 at 18:40
add a comment |
$begingroup$
Thanks. How do you reach the general term? With induction?
$endgroup$
– Sherlock Watson
Dec 19 '18 at 18:28
$begingroup$
I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
$endgroup$
– jayant98
Dec 19 '18 at 18:40
$begingroup$
Thanks. How do you reach the general term? With induction?
$endgroup$
– Sherlock Watson
Dec 19 '18 at 18:28
$begingroup$
Thanks. How do you reach the general term? With induction?
$endgroup$
– Sherlock Watson
Dec 19 '18 at 18:28
$begingroup$
I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
$endgroup$
– jayant98
Dec 19 '18 at 18:40
$begingroup$
I used these points-(1)all are odd (2)all have the difference 4 (3)2nd part in the multiplication is also differing by 4 (4) So it must be (4k+r)(4k+q) and then put the value of 1,2,.(this doesn't mean putting all the values) to get r and 1 and then checked it through other numbers. I have used these steps in many questions (never fails me).
$endgroup$
– jayant98
Dec 19 '18 at 18:40
add a comment |
$begingroup$
begin{align}
S &= sum_{n=0}^{n=22} (4n+7)(4n+11)\
&= sum_{n=0}^{n=22} 16n^2+72n+color{red}{77}\
&= 80707
end{align}
answered Dec 19 '18 at 14:40
Karn WatcharasupatKarn Watcharasupat
3,9742526
$endgroup$
add a comment |
$begingroup$
begin{align}
S &= sum_{n=0}^{n=22} (4n+7)(4n+11)\
&= sum_{n=0}^{n=22} 16n^2+72n+color{red}{77}\
&= 80707
end{align}
answered Dec 19 '18 at 14:40
Karn WatcharasupatKarn Watcharasupat
3,9742526
$endgroup$
add a comment |
$begingroup$
begin{align}
S &= sum_{n=0}^{n=22} (4n+7)(4n+11)\
&= sum_{n=0}^{n=22} 16n^2+72n+color{red}{77}\
&= 80707
end{align}
answered Dec 19 '18 at 14:40
Karn WatcharasupatKarn Watcharasupat
3,9742526
$endgroup$
begin{align}
S &= sum_{n=0}^{n=22} (4n+7)(4n+11)\
&= sum_{n=0}^{n=22} 16n^2+72n+color{red}{77}\
&= 80707
end{align}
answered Dec 19 '18 at 14:40
Karn WatcharasupatKarn Watcharasupat
3,9742526
edited Dec 19 '18 at 15:44
answered Dec 19 '18 at 14:40
Karn WatcharasupatKarn Watcharasupat
3,9742526
answered Dec 19 '18 at 14:40
Karn WatcharasupatKarn Watcharasupat
3,9742526
answered Dec 19 '18 at 14:40
Karn WatcharasupatKarn Watcharasupat
3,9742526
3,9742526
add a comment |
add a comment |
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$begingroup$
$15cdot 19=17^2-2^2$
$endgroup$
– giannispapav
Dec 19 '18 at 14:34
$begingroup$
The sequence ${9,13,17,21,ldots}$ is arithmetic with common difference $4$, so the $n$-th term is $a_n=4n+5$. Then the sum you want to evaluate is $sumlimits_{n=1}^{23}(4n+5-2)(4n+5+2)=sumlimits_{n=1}^{23}(16n^2+40n+21)$.
$endgroup$
– user170231
Dec 19 '18 at 14:43