1plz help !!
If Sn denotes the sum of the products of the first n natural numbers taken two at a time, then find the sum of the infinite series :
∞
Σ Sn(n+1)!
n=1
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12 Answers
1357e^x=1+x+x^2/2!....
differenticatin
e^x=1+2x/2!+3x^2/3!
Again differentiating
e^x=2/2!+3.2.x/3!+4.3.x^2/4!......
putting x=1 we will get the result as e
1Sn comprises of all the possible products of natural numbers taken two at a time , such that they do not repeat and it also does not include squares of natural numbers
like 1.2 + 1.3 + 2.3 + 1.4 + 2.4 + 3.4 + ........... ∞
23@ prateek
S _{n}=\frac{(\sum_{i=1}^{n}a_{i})^{2}-(\sum_{i=1}^{n}a_{i}^{2})}{2}??
1Tks qwerty i was confused whether or not 1/2 is a factor of Sn ::
Sn = 12 { n(n+1)222 - n(n+1)(2n+1)/6 }
Sn = (n+1)(n)(n-1)(3n+2)24
Now ,evaluating
∞
Σ Sn(n+1)!
n=1
= (n+1)(n)(n-1)(3n+2)/24(n+1)(n)(n-1)(n-2)!
= 1243n+2(n-2)!
= 124 { 3n-6(n-2)! + 6+2(n-2)! }
=124{ 3n-2(n-2)! + 8e }
=124 { 3(n-3)! + 8e }
=124 3e + 8e
=11e24
= ANSWER