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Expression in terms of $N$
Closed-form expression for sum of Vandermonde matrix elementsIs there a closed form expression for the Taylor series of (1- a X - b Y - c XY )^ (-1)?Closed form formula for discrete sumsHow to write a closed form expression for $sum limits_i=0^n-1a_i$ in terms of $a_0$ and $sum limits_i=1^na_i$?Binominal expression simplificationIs there a closed-form expression for $sum_k=1^nlfloor k^q rfloor$ for $q in mathbbQ_> 0$?Find $sum_k=1^n binom2n-kn(-1)^k$Closed form expression for binomial coefficientClosed form for the expansion of a polynomial expression with an arbitrary number of termsexact closed form expression
$begingroup$
Find the closed form expression in terms of $N$
$sum_t=1^N sum_i=1^t sum_k=1^i (tik)$.
Thank you for good idea.
discrete-mathematics
asked Mar 20 at 19:40
mathsstudentmathsstudent
536
$endgroup$
add a comment |
$begingroup$
Find the closed form expression in terms of $N$
$sum_t=1^N sum_i=1^t sum_k=1^i (tik)$.
Thank you for good idea.
discrete-mathematics
asked Mar 20 at 19:40
mathsstudentmathsstudent
536
$endgroup$
$begingroup$
Do you know the formula for $sum_n=1^N n$? This looks like a mild generalization of such formulas.
$endgroup$
– Clayton
Mar 20 at 19:43
$begingroup$
The formula is $n*(n+1)/2 $. Is it true ? But is not clear in this question for me.
$endgroup$
– mathsstudent
Mar 20 at 19:46
add a comment |
$begingroup$
Find the closed form expression in terms of $N$
$sum_t=1^N sum_i=1^t sum_k=1^i (tik)$.
Thank you for good idea.
discrete-mathematics
asked Mar 20 at 19:40
mathsstudentmathsstudent
536
$endgroup$
Find the closed form expression in terms of $N$
$sum_t=1^N sum_i=1^t sum_k=1^i (tik)$.
Thank you for good idea.
discrete-mathematics
discrete-mathematics
asked Mar 20 at 19:40
mathsstudentmathsstudent
536
asked Mar 20 at 19:40
mathsstudentmathsstudent
536
asked Mar 20 at 19:40
mathsstudentmathsstudent
536
asked Mar 20 at 19:40
mathsstudentmathsstudent
536
asked Mar 20 at 19:40
mathsstudentmathsstudent
536
536
$begingroup$
Do you know the formula for $sum_n=1^N n$? This looks like a mild generalization of such formulas.
$endgroup$
– Clayton
Mar 20 at 19:43
$begingroup$
The formula is $n*(n+1)/2 $. Is it true ? But is not clear in this question for me.
$endgroup$
– mathsstudent
Mar 20 at 19:46
add a comment |
$begingroup$
Do you know the formula for $sum_n=1^N n$? This looks like a mild generalization of such formulas.
$endgroup$
– Clayton
Mar 20 at 19:43
$begingroup$
The formula is $n*(n+1)/2 $. Is it true ? But is not clear in this question for me.
$endgroup$
– mathsstudent
Mar 20 at 19:46
$begingroup$
Do you know the formula for $sum_n=1^N n$? This looks like a mild generalization of such formulas.
$endgroup$
– Clayton
Mar 20 at 19:43
$begingroup$
Do you know the formula for $sum_n=1^N n$? This looks like a mild generalization of such formulas.
$endgroup$
– Clayton
Mar 20 at 19:43
$begingroup$
The formula is $n*(n+1)/2 $. Is it true ? But is not clear in this question for me.
$endgroup$
– mathsstudent
Mar 20 at 19:46
$begingroup$
The formula is $n*(n+1)/2 $. Is it true ? But is not clear in this question for me.
$endgroup$
– mathsstudent
Mar 20 at 19:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Do this in three steps:
$$
sum_k=1^i (ti) k = ti sum_k=1^i k = frac12 i^2 (1+i) t
$$
then
$$
sum_i=1^t frac12 i^2 (1+i) t = frac12 t sum_i=1^t i^2 (1+i)
= frac12 t left( sum_i=1^t (i+2) (1+i) i -sum_i=1^t 2 (1+i) iright) \
= frac12 t left( frac14 t(t+1)(t+2)(t+3) - frac23 t^2(t+1)(t+2)right) \=frac124 t^2(t+1)(t+2)(3t+1)
$$
Finally, using the same technique of breaking up into sums that all involve rising powers (of $t$ this time),
$$
frac124 t^2(t+1)(t+2)(3t+1) = frac148N^2(N+1)^2(N+2)(N+3)
$$
and that is your answer.
answered Mar 20 at 19:59
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Do this in three steps:
$$
sum_k=1^i (ti) k = ti sum_k=1^i k = frac12 i^2 (1+i) t
$$
then
$$
sum_i=1^t frac12 i^2 (1+i) t = frac12 t sum_i=1^t i^2 (1+i)
= frac12 t left( sum_i=1^t (i+2) (1+i) i -sum_i=1^t 2 (1+i) iright) \
= frac12 t left( frac14 t(t+1)(t+2)(t+3) - frac23 t^2(t+1)(t+2)right) \=frac124 t^2(t+1)(t+2)(3t+1)
$$
Finally, using the same technique of breaking up into sums that all involve rising powers (of $t$ this time),
$$
frac124 t^2(t+1)(t+2)(3t+1) = frac148N^2(N+1)^2(N+2)(N+3)
$$
and that is your answer.
answered Mar 20 at 19:59
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
$begingroup$
Do this in three steps:
$$
sum_k=1^i (ti) k = ti sum_k=1^i k = frac12 i^2 (1+i) t
$$
then
$$
sum_i=1^t frac12 i^2 (1+i) t = frac12 t sum_i=1^t i^2 (1+i)
= frac12 t left( sum_i=1^t (i+2) (1+i) i -sum_i=1^t 2 (1+i) iright) \
= frac12 t left( frac14 t(t+1)(t+2)(t+3) - frac23 t^2(t+1)(t+2)right) \=frac124 t^2(t+1)(t+2)(3t+1)
$$
Finally, using the same technique of breaking up into sums that all involve rising powers (of $t$ this time),
$$
frac124 t^2(t+1)(t+2)(3t+1) = frac148N^2(N+1)^2(N+2)(N+3)
$$
and that is your answer.
answered Mar 20 at 19:59
Mark FischlerMark Fischler
33.9k12552
$endgroup$
add a comment |
$begingroup$
Do this in three steps:
$$
sum_k=1^i (ti) k = ti sum_k=1^i k = frac12 i^2 (1+i) t
$$
then
$$
sum_i=1^t frac12 i^2 (1+i) t = frac12 t sum_i=1^t i^2 (1+i)
= frac12 t left( sum_i=1^t (i+2) (1+i) i -sum_i=1^t 2 (1+i) iright) \
= frac12 t left( frac14 t(t+1)(t+2)(t+3) - frac23 t^2(t+1)(t+2)right) \=frac124 t^2(t+1)(t+2)(3t+1)
$$
Finally, using the same technique of breaking up into sums that all involve rising powers (of $t$ this time),
$$
frac124 t^2(t+1)(t+2)(3t+1) = frac148N^2(N+1)^2(N+2)(N+3)
$$
and that is your answer.
answered Mar 20 at 19:59
Mark FischlerMark Fischler
33.9k12552
$endgroup$
Do this in three steps:
$$
sum_k=1^i (ti) k = ti sum_k=1^i k = frac12 i^2 (1+i) t
$$
then
$$
sum_i=1^t frac12 i^2 (1+i) t = frac12 t sum_i=1^t i^2 (1+i)
= frac12 t left( sum_i=1^t (i+2) (1+i) i -sum_i=1^t 2 (1+i) iright) \
= frac12 t left( frac14 t(t+1)(t+2)(t+3) - frac23 t^2(t+1)(t+2)right) \=frac124 t^2(t+1)(t+2)(3t+1)
$$
Finally, using the same technique of breaking up into sums that all involve rising powers (of $t$ this time),
$$
frac124 t^2(t+1)(t+2)(3t+1) = frac148N^2(N+1)^2(N+2)(N+3)
$$
and that is your answer.
answered Mar 20 at 19:59
Mark FischlerMark Fischler
33.9k12552
answered Mar 20 at 19:59
Mark FischlerMark Fischler
33.9k12552
answered Mar 20 at 19:59
Mark FischlerMark Fischler
33.9k12552
answered Mar 20 at 19:59
Mark FischlerMark Fischler
33.9k12552
33.9k12552
add a comment |
add a comment |
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$begingroup$
Do you know the formula for $sum_n=1^N n$? This looks like a mild generalization of such formulas.
$endgroup$
– Clayton
Mar 20 at 19:43
$begingroup$
The formula is $n*(n+1)/2 $. Is it true ? But is not clear in this question for me.
$endgroup$
– mathsstudent
Mar 20 at 19:46