Generalised Formula for Products

by SomeonecoolLovesMaths, Apr 24, 2024, 9:21 AM

$1 + 2 + 3 + … + n = \frac{n(n+1)}{2}$
Looking at the sums of the first few terms, we see that
$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + … + n(n+1)$
$= \frac{n(n+1)(n+2)}{3}$
which can easily be proved by mathematical induction, as can
$1 \cdot 2 \cdot 3 … k + 2 \cdot 3 \cdot 4 … (k+1) + … + n(n+1)(n+2)…(n+k-1)$
$= \frac{n(n+1)…(n+k)}{k+1}$
$= \frac{k!(n+k)!}{(k+1)!(n-1)!}$
$= k! \binom{n+k}{k+1}$ .

Thanks to AndrewTom

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you can also do this with a hockeystick i believe

by eg4334, Nov 17, 2024, 3:32 PM

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Can you provide your method/approach to it?

by SomeonecoolLovesMaths, Nov 17, 2024, 6:00 PM

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Here is the hockeystick proof:

$1 \cdot 2 \cdot 3 … k + 2 \cdot 3 \cdot 4 … (k+1) + … + n(n+1)(n+2)…(n+k-1)$

each term we will complete the factorial, like so (this is a very standard JEE idea):

$r(r+1)…(r+k-1) = \frac{(1)(2)\dots (r-1)r(r+1)…(r+k-1)}{(1)(2)\dots (r-1)}$

$=\frac{(r+k-1)!}{(r-1)!} = k! \frac{((r-1)+k)!}{(r-1)!k!} = k! \binom{r+k-1}{k}$
The required sum is then $k! \sum_{r=1}^n \binom{r+k-1}{k}$ which is just a hockey stick.

by quasar_lord, Mar 12, 2025, 5:06 PM

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by SomeonecoolLovesMaths, Nov 25, 2024, 5:07 PM
add this one new thing to the intergrals that is lim n tends to infinity b-a/n summation k=1 to n f(a+k(b-a)/n))= int_{a}^b f(x) dx
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Generalised Formula for Products

by SomeonecoolLovesMaths, Apr 24, 2024, 9:21 AM

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