 Sitecore Content Search is not working in the Content Editor
 Item with children got deleted from master database
 How to use Authenticated request with the Sitecore RESTful API
 What is the correct root node for 'sitecore_marketingdefinitions_master' index?
 Are Neural Net architectures accidental discoveries?
 How much does it cost to build a private Ethereum blockchain network
 Which one is a better option  web3js or golang apis?
 Function 'sendTransaction'  can any error occur before the 'transactionHash' event?
 Contract Interaction
 Command line tool to interact with smart contract
 Listening for Deposits and Transaction Confirmations Web3js
 How to use Airdrop Contract Code
 Web3j Signature and Solidity Ecrecover
 Account created using web3 is not showing in geth console
 How to check used VideoCard's resources, if good or bad?
 Israeli Highway/Road Code
 Детский вопрос: кого бьет дальнобойщик?
 Hard Lump on Finger
 Can't create a new event template from an existing event template
 ACL and Drupal roles
simple Algebraic Manipulation (factoring?)
I got stuck on some simple algebraic manipulation in a mathematical induction problem.
The statement I have is
$$\frac{k(k + 1)(2k + 1)}{6} +(k + 1)^2,$$
and I know it is equal to
$$\frac{k+1}{6}(k(2k + 1) + 6(k + 1)),$$
but I don't know the steps from one to the other, if somebody could please clarify for me.
$$\dfrac {k(k+1)(2k+1)}6+(k+1)^2=\dfrac {k(k+1)(2k+1)}6+\dfrac{6(k+1)^2}{6}=\dfrac{(k+1)[k(2k+1)+6(k+1)]}{6}$$
After factoring out $k+1$. Can you figure out the rest?
$\frac{k(k + 1)(2k + 1)}{6} +(k + 1)^2$
RESIST the urge to expand. That's a common mistake. Factoring is recognizing "chunks" and pulling them out. So there isn't any reason to "blend" before you pull out the chunks. You may have to "blend" to put the chunks in different places but you should put that off as long as you can.
So first things first. Put over common denominator.
That would be $6$.
$\frac{k(k + 1)(2k + 1)}{6} +(k + 1)^2=$
$\frac {k(k+1)(2k+1)}6 + \frac {6(k+1)

$$\dfrac {k(k+1)(2k+1)}6+(k+1)^2=\dfrac {k(k+1)(2k+1)}6+\dfrac{6(k+1)^2}{6}=\dfrac{(k+1)[k(2k+1)+6(k+1)]}{6}$$
After factoring out $k+1$. Can you figure out the rest?
20180312 21:16:39 
$\frac{k(k + 1)(2k + 1)}{6} +(k + 1)^2$
RESIST the urge to expand. That's a common mistake. Factoring is recognizing "chunks" and pulling them out. So there isn't any reason to "blend" before you pull out the chunks. You may have to "blend" to put the chunks in different places but you should put that off as long as you can.
So first things first. Put over common denominator.
That would be $6$.
$\frac{k(k + 1)(2k + 1)}{6} +(k + 1)^2=$
$\frac {k(k+1)(2k+1)}6 + \frac {6(k+1)^2}6=$
$\frac {k(k+1)(2k+1) + 6(k+1)^2}6$.
Now look for common "chunks". There is a $(k+1)$ that both terms in the numerator have. Pull it out.
$\frac {(k+1)[k(2k+1) + 6(k+1)]}6$
20180312 21:22:23