# ISI2014-DCG-23

185 views

The sum of the series $\:3+11+\dots +(8n-5)\:$ is

1. $4n^2-n$
2. $8n^2+3n$
3. $4n^2+4n-5$
4. $4n^2+2$

recategorized

Answer$\mathbf A$

Explanation:

The above series forms an $\mathbf {AP}$

where, $\mathrm {a = 3, \;d = 8,\;a_n = (8n-5)}$

So, sum of n terms of $\mathbf {AP}$ is given by:

\begin {align} \mathrm {S_n} &= \mathrm {\frac{n}{2}[a + a_n] }\\&= \mathrm {\frac{n}{2}[3 + 8n-5] }\\&=\mathrm { \frac{n}{2}[8n-2] }\\&= \mathrm {n(4n-1)}\end {align}

$\therefore \mathrm{S_n = 4n^2-n}$

$\therefore \mathbf A$ is the right option.

edited by

Jeet's answer is good enough, but if you are looking for another approach, it could be :

3 + 11 + ... + (8n-5) = $\sum (8n-5) = 8n(n+1)/2 - 5n = 4n^2 - n$

The summation function is $n(n+1)/2$ instead of $n(n-1)/2$ because in summation, the starting value of n is taken as 0, whereas, we want it to start from 1 here.

## Related questions

1
158 views
If $l=1+a+a^2+ \dots$, $m=1+b+b^2+ \dots$, and $n=1+c+c^2+ \dots$, where $\mid a \mid <1, \: \mid b \mid < 1, \: \mid c \mid <1$ and $a,b,c$ are in arithmetic progression, then $l, m, n$ are in arithmetic progression geometric progression harmonic progression none of these
If the sum of the first $n$ terms of an arithmetic progression is $cn^2$, then the sum of squares of these $n$ terms is $\frac{n(4n^2-1)c^2}{6}$ $\frac{n(4n^2+1)c^2}{3}$ $\frac{n(4n^2-1)c^2}{3}$ $\frac{n(4n^2+1)c^2}{6}$
The number of divisors of $6000$, where $1$ and $6000$ are also considered as divisors of $6000$ is $40$ $50$ $60$ $30$
The sum of the series $\dfrac{1}{1.2} + \dfrac{1}{2.3}+ \cdots + \dfrac{1}{n(n+1)} + \cdots$ is $1$ $1/2$ $0$ non-existent