Question

I am not able to arrive at the same first part after applying theorems

Answer 1

\[\sum_{i=1}^{n}\left( 2+\frac{ i }{ n } \right)\left( \frac{ 1 }{ n } \right)\]

Answer 2

I keep getting
(4n^2 + 4ni/n^2 + i^2/n^2) (1/n)

Answer 3

2/n + i/n^2

2/n is constant

in i/n^2, 1/n^2 is constant

Answer 4

i believe you know whats summation of constant from 1 to n

and whats summation of i from i = 1 to n
?

Answer 5

\(\sum \limits_{i=1}^n c = n \\ \sum \limits_{i=1}^n n = \dfrac{n(n+1)}{2}\)

Answer 6

2/n times n + (1/n^2) times (n) (n+1)/2
simplify

Answer 7

I get those two theorems
sorry I forgot the squared, but when I apply
\[\sum_{i=1}^{n}\left( 2n+\frac{ i }{ n } \right)^{2}\left( \frac{ 1 }{ n } \right)\]

Answer 8

ok, so u'll need one more formula
\(\sum \limits_{i=1}^n i^2 = \dfrac{n (n+1)(2n+1)}{6}\)


2n + 4 n (n+1)/2 + (1/n^3) times n (n+1)(2n+1)/6

Answer 9

u get how i got that last expression ?
(upon solving summation, you would surely not get any term in "i" )

Answer 10

that part I understand
the book is telling me that it is
(4n^2+4ni+i^2)/n^2 (1/n)

Answer 11

ofcourse
thats without applying summation

(a+b)^2 = a^2+2ab+b^2
(2n+i/n)^2 =....

Answer 12

I am not getting the 4n^2/n^2 part just 4n^2

Answer 13

\(\large \left( 2n+\frac{ i }{ n } \right)^{2}\left( \frac{ 1 }{ n } \right) = (4n^2 +2 \times 2n \times (i/n) +(i/n)^2) \: (1/n) \\ \large = (4n^2 +4i +i^2/n^2) (1/n) \)

Answer 14

something's not correct....
either question or answer in your book

Answer 15

you and I evaluated the same way, it's the book that is throwing me off

Answer 16

your book did
\((2+i/n)^2\)

Answer 17

so whats the question ??

because at first you posted
(2+i/n)

then you said you missed the square sign

and posted
(2n+i/n)^2
(instead of (2+i/n)^2)

Answer 18

if it is actually
\((2+i/n)^2\)
then your book is correct

Answer 19

the original question was
(2+i/n)^2 (1/n)

Answer 20

aah, then what you posted was not correct.
your book is correct

[(2n+i)/n ]^2

Answer 21

I thought I should evaluate that squared first, then apply the theorem

Answer 22

I see what you did there, that was my suspicion

Answer 23

thanks a lot, I was stuck all morning trying to rework this one

Answer 24

welcome ^_^