Question

Please find attached

Answer 1

ack bad camera.. .its not visible properly lol

Answer 2

its "11" above ?

Answer 3

over SUM

Answer 4

n above the sigm i=1 below it.

Answer 5

You can write it as:

\[\large \sum_{i=1}^{n}(4i + 2) \implies \sum_{i=1}^{n}(4i) + \sum_{i=1}^{n}(2)\]

Answer 6

Or:

\[4\sum_{i=1}^{n}i + 2n \implies 4 (\frac{n(n+1)}{2}) + 2n\]

Answer 7

sorry that confused me.

Answer 8

Using the formula:

\[\large \sum_{k= 1}^{n}(k) = \frac{n(n+1)}{2}\]

Answer 9

Where you did not get??

Answer 10

no no no, I get it.

Answer 11

See,

\[\sum_{k = 1}^{n} = 1 + 2 + 3+ 4............+n = \frac{n(n+1)}{2}\]

Answer 12

And adding 2 n times you will get 2n..

Answer 13

2n^2+4n???

Answer 14

\[\implies 2n(n+1) + 2n = 2n^2 + 4n\]

Yes you are absolutely right..