Question

How do you find the sum of $\sum{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}$ where i is [1, 4]?

Answer 1

Hint: We will consider $\sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}$ and apply algebraic formulas ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$ to solve it further. We are going to apply the formulas separately to ${{\left( i-1 \right)}^{2}}$ and \[{{\left( i+1 \right)}^{3}}\]. After getting the resultant terms we will simply substitute them in $\sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}$. Finally we will put I as 1, 2, 3, 4 and solve it in the required correct way to get the right sum.

Complete step by step solution:
The right way to understand the question is to write it as $\sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}$ …(i).
We did this because it is given in the question that i is [1, 4]. Here, we will apply two algebraic equations and these are ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$. We will apply ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to ${{\left( i-1 \right)}^{2}}$ thus, we get
$\begin{align}
  & {{\left( i-1 \right)}^{2}}={{\left( i \right)}^{2}}+{{\left( 1 \right)}^{2}}-2\left( i \right)\left( 1 \right) \\
 & \Rightarrow {{\left( i-1 \right)}^{2}}={{i}^{2}}+1-2i\,\,\,\,...(ii) \\
\end{align}$
Similarly, we will apply ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$ to \[{{\left( i+1 \right)}^{3}}\] and get the following.
\[\begin{align}
  & {{\left( i+1 \right)}^{3}}={{\left( i \right)}^{3}}+{{\left( 1 \right)}^{3}}+3{{\left( 1 \right)}^{2}}\left( i \right)+3\left( 1 \right){{\left( i \right)}^{2}} \\
 & \Rightarrow {{\left( i+1 \right)}^{3}}={{i}^{3}}+1+3i+3{{i}^{2}}\,\,\,...(iii) \\
\end{align}\]
Now, we will substitute (ii) and (iii) in (i) to get the following.
 \[\begin{align}
  & \sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}=\sum\limits_{i=1}^{4}{\left[ \left( {{i}^{2}}+1-2i \right)+\left( {{i}^{3}}+1+3i+3{{i}^{2}} \right) \right]} \\
 & \Rightarrow \sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}=\sum\limits_{i=1}^{4}{\left[ {{i}^{2}}+1-2i+{{i}^{3}}+1+3i+3{{i}^{2}} \right]} \\
 & \Rightarrow \sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}=\sum\limits_{i=1}^{4}{\left[ {{i}^{3}}+4{{i}^{2}}+i+2 \right]} \\
 & \Rightarrow \sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}=\left[ {{\left( 1 \right)}^{3}}+4{{\left( 1 \right)}^{2}}+\left( 1 \right)+2 \right]+\left[ {{\left( 2 \right)}^{3}}+4{{\left( 2 \right)}^{2}}+\left( 2 \right)+2 \right] \\
 & +\left[ {{\left( 3 \right)}^{3}}+4{{\left( 3 \right)}^{2}}+\left( 3 \right)+2 \right]+\left[ {{\left( 4 \right)}^{3}}+4{{\left( 4 \right)}^{2}}+\left( 4 \right)+2 \right] \\
 & \Rightarrow \sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}=\left[ 1+4+1+2 \right]+\left[ 8+16+2+2 \right]+\left[ 27+36+3+2 \right]+\left[ 64+64+4+2 \right] \\
 & \Rightarrow \sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}=8+28+68+134 \\
 & \Rightarrow \sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}=238 \\
\end{align}\]

Hence, the required sum is 238.

Note: We could have solved it by starting to open the sum first. This is done below.
\[\begin{align}
  & \sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}=\left[ {{\left( 1-1 \right)}^{2}}+{{\left( 1+1 \right)}^{3}} \right]+\left[ {{\left( 2-1 \right)}^{2}}+{{\left( 2+1 \right)}^{3}} \right]+\left[ {{\left( 3-1 \right)}^{2}}+{{\left( 3+1 \right)}^{3}} \right]+\left[ {{\left( 4-1 \right)}^{2}}+{{\left( 4+1 \right)}^{3}} \right] \\
 & \Rightarrow \sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}=\left[ 0+{{\left( 2 \right)}^{3}} \right]+\left[ {{\left( 1 \right)}^{2}}+{{\left( 3 \right)}^{3}} \right]+\left[ {{\left( 2 \right)}^{2}}+{{\left( 4 \right)}^{3}} \right]+\left[ {{\left( 3 \right)}^{2}}+{{\left( 5 \right)}^{3}} \right] \\
 & \Rightarrow \sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}=8+1+27+4+64+9+125 \\
 & \Rightarrow \sum\limits_{i=1}^{4}{\left[ {{\left( i-1 \right)}^{2}}+{{\left( i+1 \right)}^{3}} \right]}=238 \\
\end{align}\]
We can clearly see that we did not even apply algebraic formulas for a solution. This is a simple method to get the right answer in no time. The above is a little lengthy and this is why it needs focus while solving it. In case we lose our concentration then we might lead towards the wrong answer. Do not misunderstand I as iota otherwise, we will not get the required answer which is what we need here. Take care of minus and plus signs while using algebraic formulas.