Question

The ratio of the sums of first $n$ even numbers and $n$ odd numbers will be.

• A. $1 : n$
• B. $( n + 1 ) : 1$
• C. $( n + 1 ) : n$
• D. $( n - 1 ) : 1$

Answer 1

Answer: (C)

$( n + 1 ) : n$

Answer 2

Let $S _ { \text {Even } } = 2 + 4 + 6 + 8 + \ldots \ldots . . \infty$ …..(i)

And $S _ { \text {Odd } } = 1 + 3 + 5 + 7 + 9 + \ldots \ldots \infty$ …..(ii)

Sum $S _ { E } = \frac { n } { 2 } [ 4 + ( n - 1 ) 2 ] = \frac { n } { 2 } [ 2 n + 2 ] = \frac { n } { 2 } 2 ( n + 1 )$

And $S _ { O } = \frac { n } { 2 } [ 2 + ( n - 1 ) 2 ] = \frac { n } { 2 } ( 2 n )$

Now$\frac { S _ { E } } { S _ { O } } = \frac { ( n + 1 ) } { n }$ or $S _ { E } : S _ { O } = ( n + 1 ) : n$.