The value of 23!+45!+67!+..... is
e12
e-1
e
e-13
Explanation for the correct answer:
Let the given summation be S=23!+45!+67!+.....
The nth term in this summation can be written as
tn=2n2n+1!
⇒ tn=2n+1-12n+1!
⇒ tn=2n+12n+1!-12n+1!
⇒ tn=12n!-12n+1!
⇒ t1=12!-13!
⇒ t2=14!-15!
∴ S=∑tn
⇒ S=12!-13!+14!-15!+16!-17!+....
⇒ S=1-1+12!-13!+14!-15!+16!-17!+....
We know that, ex=1+x1!+x22!+x33!+....
Substitute x=-1
e-1=1-1+12!-13!+14!+....
⇒ S=e-1
Hence, the value of 23!+45!+67!+..... is e-1.
Hence, option B is the correct answer.
The sum of the series 23!+45!+67!+... to ∞ = ae. Find (a+3)2.
If f=x1+x2+13(x1+x2)3+15(x1+x2)5+... to ∞ and g=x−23x3+15x5+17x7−29x9+..., then f=d×g. Find 4d.