In the inductive step, it needs to prove that, if\(P\left( k \right)\)is true, then\(P\left( {k + 1} \right)\)is also true.
That is,\(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.
In the inductive hypothesis, it assumes that\(P\left( k \right)\)is true for any arbitrary positive integer\(k\).
That is, it is written as\(\sum\limits_{j = 1}^k {{j^2}} {2^j} = {k^2}{2^{k + 1}} - k{2^{k + 2}} + 3 \cdot {2^{k + 1}} - 6\).
Now it must have to show that\(P\left( {k + 1} \right)\)is also true.
\(\sum\limits_{j = 1}^{k + 1} {{j^2}} {2^j} = \sum\limits_{j = 1}^k {{j^2}} {2^j} + {\left( {k + 1} \right)^2}{2^{k + 1}}\)
From inductive hypothesis it knows \(\sum\limits_{j = 1}^k {{j^2}} {2^j} = {k^2}{2^{k + 1}} - k{2^{k + 2}} + 3 \cdot {2^{k + 1}} - 6\).
Therefore, it is written as:
\(\begin{aligned}{c}\sum\limits_{j = 1}^{k + 1} {{j^2}} {2^j} = {k^2}{2^{k + 1}} - k{2^{k + 2}} + 3 \cdot {2^{k + 1}} - 6 + {\left( {k + 1} \right)^2}{2^{k + 1}} - 6\\ &= {2^{k + 1}}\left( {{k^2} - 2k + 3 + {{\left( {k + 1} \right)}^2}} \right) - 6\\ &= {2^{k + 1}}\left( {{k^2} - 2k + 3 + {k^2} + 2k + 1} \right) - 6\\ &= {2^{k + 1}}\left( {2{k^2} + 4} \right) - 6\end{aligned}\)
Further, solve the above expression,
\(\begin{aligned}{c}\sum\limits_{j = 1}^{k + 1} {{j^2}} {2^j} &= {2^{k + 2}}\left( {{k^2} + 2} \right) - 6\\ &= {2^{k + 2}}\left( {{k^2} + 2k + 1 - 2k - 2 + 3} \right) - 6\\ &= {2^{k + 2}}\left( {{{\left( {k + 1} \right)}^2} - 2\left( {k + 1} \right) + 3} \right) - 6\\ &= {\left( {k + 1} \right)^2}{2^{k + 2}} - \left( {k + 1} \right){2^{k + 3}} + 3 \cdot {2^{k + 2}} - 6\end{aligned}\)
From the above, it can see that\(P\left( {k + 1} \right)\)is also true.
Hence,\(P\left( {k + 1} \right)\)is true under the assumption that\(P\left( k \right)\)is true. This
completes the inductive step.
Hence, using mathematical induction, it is proved that\(\sum\limits_{j = 1}^n {{j^2}} {2^j} = {n^2}{2^{n + 1}} - n{2^{n + 2}} + 3 \cdot {2^{n + 1}} - 6\)for every positive integer n.
