A) \[1:n\]
B) \[(n+1):1\]
C) \[(n+1):n\]
D) \[(n-1):1\]
Correct Answer: C
Solution :
Let \[{{S}_{Even}}=2+4+6+8+........\infty \] ?..(i) and \[{{S}_{Odd}}=1+3+5+7+9+.......\infty \] ?..(ii) Sum \[c\] and \[{{S}_{O}}=\frac{n}{2}[2+(n-1)2]=\frac{n}{2}(2n)\] Now \[\frac{{{S}_{E}}}{{{S}_{O}}}=\frac{(n+1)}{n}\] or \[{{S}_{E}}:{{S}_{O}}=(n+1):n\].
You need to login to perform this action.
You will be redirected in
3 sec
