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8. Sequences and Series
easy
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
$(n - 1):1$
Solution
(c) Let ${S_{Even}} = 2 + 4 + 6 + 8 + ……..\infty $ …..$(i)$
and ${S_{Odd}} = 1 + 3 + 5 + 7 + 9 + …….\infty $ …..$(ii)$
${S_E} = \frac{n}{2}[4 + (n – 1)2] = \frac{n}{2}(2n + 2) = n(n + 1)$
${S_O} = \frac{n}{2}[2 + (n – 1)2] = \frac{n}{2}(2n)$
Now $\frac{{{S_E}}}{{{S_O}}} = \frac{{(n + 1)}}{n}$
or ${S_E}:{S_O} = (n + 1):n$.
Standard 11
Mathematics