jee advanced

[kindred heron]

Solution

[pulsar temple]

what does this mean

gamma function of 1.0, 3.1, 7.2?

[kindred heron]

It's factorial ...
Factorial of sequnce is equal to 4000

Example 3! Is 3×2×1 = 6

[azure coral]

i DESPISE when they use . to mean multiplication

[azure coral]

it's $1\cdot 0! + 3\cdot 1! + 7\cdot 2!$

[drifting fox]

artemetra

[azure coral]

it makes me irrationally angry lol

[hollow ore]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[azure coral]

try writing the general formula for the kth term in this sum

[autumn ore]

3999?

just verifying the asnwer

[gray flame]

isnt this a sequence?

Numbers before “.” Are coefficients

Like

1. Coefficient = 1 at 0!

And so on

Look at the coefficients 1, 3, 7, 13, 21:

Calculate the differences:
• 3 - 1 = 2
• 7 - 3 = 4
• 13 - 7 = 6
• 21 - 13 = 8

The first difference sequence is 2, 4, 6, 8, which is arithmetic, so the original coefficients follow the pattern:

$a_k = k^2 + k + 1$

[drifting fox]

i'mnoTx!ter

[gray flame]

Now

Verify for K

i am too lazy

S = (n+2)! - (n+1)!

Step 2: Write the sum explicitly

Sum S is:

[drifting fox]

i'mnoTx!ter

[gray flame]

Step 3: Simplify the sum

Rewrite the coefficient as:

$k^2 + k + 1 = k(k + 1) + 1$

[drifting fox]

i'mnoTx!ter

[gray flame]

So,

$S = \sum_{k=0}^n \left[k(k+1) + 1\right] k! = \sum_{k=0}^n k(k+1)k! + \sum_{k=0}^n k!$

[drifting fox]

i'mnoTx!ter

[gray flame]

Step 4: Simplify the first sum

Note that:

$k(k+1)k! = k(k+1)k! = (k^2 + k)k!$

[drifting fox]

i'mnoTx!ter

[gray flame]

But we want to rewrite k(k+1)k! in factorial form.

Observe:

$(k+2)! = (k+2)(k+1)k!$

[drifting fox]

i'mnoTx!ter

[gray flame]

We can express k(k+1)k! in terms of factorials.

Try expressing k(k+1)k! in terms of factorials:

$k(k+1)k! = (k+1)k \cdot k! = (k+1) \cdot k \cdot k! = (k+1) \cdot (k!)
\cdot k$

[drifting fox]

i'mnoTx!ter

[gray flame]

Not very direct. Instead, use the identity:

$(k+2)! = (k+2)(k+1)k! = (k^2 + 3k + 2)k!$

[drifting fox]

i'mnoTx!ter

[gray flame]

Compare k(k+1)k! = (k^2 + k)k! to (k+2)! = (k^2 + 3k + 2)k!

So,

k(k+1)k! = (k+2)! - 2(k+1)k! = (k+2)! - 2(k+1)!

its too long

I amma js write steps

Step 5: Substitute back in the sum

Step 6: Separate the sum

Step 7: Rewrite the sums with index shifts

Step 8: Combine and simplify sums

Step 9: Group sums

Step 10: Write out sums with last terms separated

Step 11: Final simplified sum
Step 12: Use

$S = (n+2)! - (n+1)!$

[drifting fox]

i'mnoTx!ter

[gray flame]

Yes

[carmine pike]

What else could it mean?

[azure coral]

decimal point

(3.1)!

[carmine pike]

How is factorial defined for decimal numbers?

[azure coral]

using gamma function

[carmine pike]

Im pretty sure this is pre uni math

[azure coral]

in my defence india is crazy with math so i wouldn't be surprised if they did something like that lol

[carmine pike]

I get you but jee advanced doesn't go that far beyond

[azure coral]

in any case explicit > implicit

[kindred heron]

Yeha it is

[kindred heron]

Thanks for this solution

.solved