Question

Solve the given expression:

$\sum\limits_{k=1}^{k=n}{k\left( k+1 \right)\left( k+2 \right)}$

Answer 1

Hint: Express the general term (kth term) of the given series i.e. k (k + 1) (k + 2) as
$k\left( k+1 \right)\left( k+2 \right)=\dfrac{1}{4}\left[ k\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)-\left( k-1 \right)k\left( k+1 \right)\left( k+2 \right) \right]$
Now, write the terms of series by putting k = 1, 2, 3, 4………….n and observe the series formed by the above equation, cancel out the same terms with negative and positive signs and hence get summation.

Complete step-by-step answer:
Expression given in the problem is
$\sum\limits_{k=1}^{k=n}{k\left( k+1 \right)\left( k+2 \right)}....................\left( i \right)$
Let us suppose the sum represented in equation (i) is ‘s’. So, we get
$s=\sum\limits_{k=1}^{k=n}{k\left( k+1 \right)\left( k+2 \right)}..................\left( ii \right)$
Now, let us represent the general term (kth term) of the given series as ${{T}_{k}}$ . So, we get
${{T}_{k}}=k\left( k+1 \right)\left( k+2 \right).........................\left( iii \right)$
Now, we can write the above equation by creating the equation in different forms. So, let us observe the equation
k (k + 1) (k + 2) (k + 3) – (k – 1) k(k + 1) (k + 2)…………………(iv)
So, take k (k + 1) (k + 2) as common from the above equation. So, we get
= k (k + 1) (k + 2) [(k + 3) – (k – 1)]
= k (k + 1) (k + 2) (k + 3 – k + 1)
= 4k (k + 1) (k + 2)…………………………(v)
Hence, we can represent 4k (k +1) (k + 2) as
4k (k + 1) (k + 2) = k (k + 1) (k + 2) (k + 3) – (k -1) (k) (k + 1) (k + 2) or
$k\left( k+1 \right)\left( k+2 \right)=\dfrac{1}{4}\left[ k\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)-\left( k-1 \right)\left( k \right)\left( k+1 \right)\left( k+2 \right) \right].....................\left( vi \right)$
So, we can represent ${{T}_{k}}$ from the equation (iii) with the help of equation (vi) as
${{T}_{k}}=\dfrac{1}{4}\left[ k\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)-\left( k-1 \right)k\left( k+1 \right)\left( k+2 \right) \right].....................\left( vii \right)$
Hence, we can get sum ‘s’ by putting k = 1, 2, 3………n as equation (ii) has the same general term as given in equation (vii). So, we get sum ‘s’ as
$s=\dfrac{1}{4}\sum\limits_{k=1}^{n}{\left[ k\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)-\left( k-1 \right)k\left( k+1 \right)\left( k+2 \right) \right]}$
Now, we can get s by putting k = 1, 2, 3……………..n. So, we can write ‘s’ as
$s=\dfrac{1}{4}\left[ \begin{align}
  & 1.2.3.4-0.1.2.3+ \\
 & 2.3.4.5-1.2.3.4+ \\
 & 3.4.5.6-2.3.4.5+ \\
 & 4.5.6.7-3.4.5.6+ \\
 & .\text{ }\text{.} \\
 & .\text{ }\text{.} \\
 & .\text{ }\text{.} \\
 & .\text{ }\text{.} \\
 & \left( n-1 \right)n\left( n+1 \right)\left( n+2 \right)-\left( n-2 \right)\left( n-1 \right)n\left( n+1 \right)+n\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)-\left( -1 \right)n\left( n+1 \right)\left( n+2 \right) \\
\end{align} \right]$
Now, we can observe the above expression written in the form of
${{T}_{1}}+{{T}_{2}}+{{T}_{3}}+{{T}_{4}}+...................+{{T}_{n}}$
Sum of the second third term till last term we can cancel out 1.2.3.4 and -1.2.3.4 from first row and second row similarly, we can cancel out 2.3.4.5 and -2.3.4.5 from second and third row respectively and similarly apply the same process up to the last row. Hence, we will get only one term remaining with the series, given as
$s=\dfrac{1}{4}\left[ n\left( n+1 \right)\left( n+2 \right)\left( n+3 \right) \right]$
Hence, sum of the given series in the problem is calculated as
$s=\dfrac{1}{4}\left[ n\left( n+1 \right)\left( n+2 \right)\left( n+3 \right) \right]$

Note: Another approach for the given question would be given as
$\begin{align}
  & =\sum{k\left( k+1 \right)\left( k+2 \right)} \\
 & =\sum{\left( {{k}^{2}}+k \right)\left( k+2 \right)=\sum{\left( {{k}^{{3}}}+{{k}^{2}}+2{{k}^{2}}+2k \right)}} \\
 & =\sum{\left( {{k}^{3}}+3{{k}^{2}}+2k \right)} \\
 & =\sum{{{k}^{3}}+3\sum{{{k}^{2}}+2\sum{k}}} \\
 & =\left( \dfrac{n{{\left( n+1 \right)}^{2}}}{2} \right)+3\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{n\left( n+1 \right)}{2} \\
\end{align}$
Now, solve the above expression to get the answer. Writing k (k + 1) (k + 2) in different forms is the key point of these kinds of problems. One may apply the same approach with these kinds of general terms. Example:
$\begin{align}
  & k\left( k+1 \right)=\dfrac{1}{3}\left[ k\left( k+1 \right)\left( k+2 \right)-\left( k-1 \right)k\left( k+1 \right) \right] \\
 & k\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)=\dfrac{1}{5}\left[ k\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)\left( k+4 \right)-\left( k-1 \right)k\left( k+1 \right)\left( k+2 \right)\left( k+3 \right) \right] \\
\end{align}$
We can apply the same approach if the general term consist of the same terms in fraction as
$\begin{align}
  & \dfrac{1}{k\left( k+1 \right)}=\left[ \dfrac{1}{k}-\dfrac{1}{k+1} \right] \\
 & \dfrac{1}{k\left( k+1 \right)\left( k+2 \right)}=\dfrac{1}{2}\left[ \dfrac{1}{k\left( k+1 \right)}-\dfrac{1}{\left( k+1 \right)\left( k+2 \right)} \right] \\
 & \dfrac{1}{k\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)}=\dfrac{1}{3}\left[ \dfrac{1}{k\left( k+1 \right)\left( k+2 \right)}-\dfrac{1}{\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)} \right] \\
\end{align}$