Question

$\huge\tt\pink{ \gamma | Question:- | \gamma }\ \textless \ br /\ \textgreater$
$1 + \frac{ {1}^{2} + {1}^{2} }{2!} + \frac{ {1}^{2} + {2}^{2} + {3}^{2} }{3!} + \frac{ {1}^{2} + {2}^{2} + {3}^{2} + {4}^{2} }{4!} + \: ... \infty = \frac{b}{6} e. \: Find \: b$
Solve step by step very easy question for that people who know this . ✔️​

Answer 1

Answer:

Sorry but just a little bit of correction, in the second term it should have been 1^2+2^2/2! instead of 1^2+1^2/2! but no problem tn was still identifiable seeing the other terms ;).

Prerequisites that I'll use

Summation from r=0 to infinity of 1/r! = e

(where e=2.7182818284590452353..... indeed an irrational number)

tn = summation r^2 from r=1 to n / n!

So,

tn = n(n+1)(2n+1)/6n! = (n+1)(2n+1)/6(n-1)!

So, tn = 2n^2+3n+1/6(n-1)!

tn = 1/6((n-1)+ 2n^2 +2n+2 / n-1! )

= 1/6 ( 1/n-2! + 2n^2 + 2n + 2/n-1!)

= 1/6 ( 1/n-2! + (n-1)+2n^2 + n+3 /n-1!

= 1/6(2/n-2! + (2n^2+n+3)/n-1! )

= 1/6(2/n-2! + (n-1)+(2n^2 +4)/n-1!)

= 1/6(3/n-2! + 2n^2+4/n-1! )

= 1/6(3/n-2! + 2n^-2+6/n-1!)

= 1/6(3/n-2! + 2(n-1)(n+1)/n-1! + 6/n-1!)

= 1/6(3/n-2! + 2(n+1)/n-2! + 6/n-1!)

= 1/6(3/n-2! + 2(n-2+3)/n-2! + 6/n-1!)

= 1/6(3/n-2! + 2/(n-3)! +6/n-2! + 6/n-1!)

= 1/6(2/n-3! + 9/n-2! + 6/n-1!)

Now, after this, we know that the summation runs from n=0 to infinity

but this means that there are going to be terms of (-3)!, (-2)! and (-1)! in the denominator but we know that all of those terms are going to be zero in the limit as by Gamma function definition factorial of all the negative integers are going to tend to plus or minus infinity.

This means after applying the summation the answer will straight away be

S = 1/6(2e+9e+6e) = 17e/6

and this means finally b=17 according to the answer asked.

Answer 2

Step-by-step explanation:

Answer ✍️

____________

• Sorry but just a little bit of correction, in the second term it should have been 1^2+2^2/2! instead of 1^2+1^2/2! but no problem tn was still identifiable seeing the other terms ;).

Prerequisites that I'll use

Summation from r=0 to infinity of 1/r! = e

• (where e=2.7182818284590452353. indeed an irrational number)

tn = summation r^2 from r=1 to n / n!

So,

tn = n(n+1)(2n+1)/6n! = (n+1)(2n+1)/6(n-1)!

So, tn = 2n^2+3n+1/6(n-1)!

tn = 1/6((n-1)+ 2n^2 +2n+2 / n-1! )

= 1/6 ( 1/n-2! + 2n^2 + 2n + 2/n-1!)

= 1/6 ( 1/n-2! + (n-1)+2n^2 + n+3 /n-1!

= 1/6(2/n-2! + (2n^2+n+3)/n-1! )

= 1/6(2/n-2! + (n-1)+(2n^2 +4)/n-1!)

= 1/6(3/n-2! + 2n^2+4/n-1! )

= 1/6(3/n-2! + 2n^-2+6/n-1!)

= 1/6(3/n-2! + 2(n-1)(n+1)/n-1! + 6/n-1!)

= 1/6(3/n-2! + 2(n+1)/n-2! + 6/n-1!)

= 1/6(3/n-2! + 2(n-2+3)/n-2! + 6/n-1!)

= 1/6(3/n-2! + 2/(n-3)! +6/n-2! + 6/n-1!)

= 1/6(2/n-3! + 9/n-2! + 6/n-1!)

Now, after this, we know that the summation runs from n=0 to infinity

• but this means that there are going to be terms of (-3)!, (-2)! and (-1)! in the denominator but we know that all of those terms are going to be zero in the limit as by Gamma function definition factorial of all the negative integers are going to tend to plus or minus infinity.

This means after applying the summation the answer will straight away be

S = 1/6(2e+9e+6e) = 17e/6

and this means finally b=17 according to the answer asked.

hope this helps you