Question

s1- 2+4+.......+2n and s2-1+3+...+(2n-1) then s1:s2
a . n+1/2
b.n/n+1
c.n(n)
d.n+1

Answer 1

Given,

Two series are given;

S1 - 2+4+ up to +2n

S2 - 1+3+ up to +(2n-1)

To find,

The value of S1:S2.

Solution,

We can simply solve this mathematical problem using the following process:

Mathematically,

The sum of n terms of the series of Furst n natural numbers is equal to n(n+1)/2.

Now, according to the question;

S1 = 2+4+ up to +2n

= 2 x (1+2+3+ up to +n)

= 2 x n(n+1)/2

= n(n+1)

=> S1 = n(n+1) {Equation-1}

And,

S2 = 1+3+...+(2n-1)

= (2-1) + (4-1) + (6-1) + up to + (2n-1) {n terms}

= (2+4+6+ up to +2n) - (1+1+ up to +1) {n terms in each series}

= 2 x (1+2+ up to +n) - n x 1

= 2 x n(n+1)/2 - n

= n(n+1) - n

= n (n+1-1)

= n^2

=> S2 = n^2 {Equation-2}

Now,

the value of S1:S2

= n(n+1) / n^2 {by using equations 1 and 2}

= (n+1)/n

=> S1:S2::(n+1):n

Hence, the value of S1:S2 is equal to (n+1)/n.

Answer 2

Therefore,

The Answer to this question will be,

$\frac{n + 1}{n}$