Question

1. Find
the
Sum :

1/2, 2/3 and 3/4​

Answer 1

Answer:

The sum of the finite series

12!+23!+34!+…k(k+1)! 12!+23!+34!+…k(k+1)!  

is equal to

(k+1)!−1(k+1)!. (k+1)!−1(k+1)!.  

One way to show this is to consider the partial sums Sn Sn , that is, the sum of the first n n terms:

S1=1/2 S1=1/2  

S2=1/2+2/6=5/6 S2=1/2+2/6=5/6  

S3=5/6+3/24=23/24 S3=5/6+3/24=23/24  

and from these, notice the pattern that the denominator is always equal to (n+1)! (n+1)! and the numerator is one less than the denominator, (n+1)!−1 (n+1)!−1  . Thus the part sum Sn=(n+1)!−1(n+1)! Sn=(n+1)!−1(n+1)!  

We can prove that this is true by induction.The base case is to show that this works for n=1 n=1 :

S1=2−12!=1/2 S1=2−12!=1/2  

and that is certainly true.

Now the inductive hypothesis states that, if Sn=(n+1)!−1(n+1)! Sn=(n+1)!−1(n+1)! then Sn+1=(n+2)!−1(n+2)! Sn+1=(n+2)!−1(n+2)!  

Let's see if this holds. Note that the n+1 partial sum is equal to the n-th partial sum, plus the n+1 term:

Sn+1=Sn+n+1(n+2)! Sn+1=Sn+n+1(n+2)!  

Sn+1=(n+1)!−1(n+1)!+n+1(n+2)! Sn+1=(n+1)!−1(n+1)!+n+1(n+2)!  

We find a common denominator, multiplying the first term's numerator and denominator by (n+2) (n+2) :

Sn+1=(n+2)(n+1)!−(n+2)+n+1(n+2)!=(n+2)!−1(n+2)! Sn+1=(n+2)(n+1)!−(n+2)+n+1(n+2)!=(n+2)!−1(n+2)!  

Meaning that our inductive hypothesis is true. Finally, we plug in n=k n=k  to obtain the value of the finite sum:

Sk=(k+1)!−1(k+1)!

Step-by-step explanation:

Answer 2

Answer:

$1 \div 2 + 2 \div 3 + 3 \div 4$

L.C.M. of 2,3 and 4 = 12

= (6+8+9)/12

= 23/12

= 1.91

Step-by-step explanation:

Thank u