The sum of the infinite series 1+12!+1·34!+1·3·56!+… is equal to
e
e2
e12
1e
Explanation for the correct option:
Given, 1+12!+1·34!+1·3·56!+…
Now, nth term of this series=1·3·...·(2n-1)2n! (excluding first term)
⇒Tn=1·3·...·(2n-1)2n!×(2·4·...·2n)(2·4·...·2n)⇒Tn=2n!2n!×2nn!=12nn!
So,
⇒Sn=1+12×1!+122×2!+123×3!+...
We know that,
ex=1+x1!+x22!+x33!+...
When x=12,
e12=1+121!+1222!+1233!+...=1+12×1!+122×2!+123×3!+...=Sn
∴Sn=e12
Hence, option C is correct.
The sum of the infinite series 1+23+732+1233+1734+2235+…. is equal to
The product of the following series (1+11!+12!+13!+...) × (1−11!+12!−13!+...) is