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Question
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
Solution
Let P(n) be the given statement.
Now,
\[P(n) = 1 . 2 + 2 . 2^2 + 3 . 2^3 + . . . + n . 2^n = (n - 1) 2^{n + 1} + 2\]
\[\text{ Step 1} : \]
\[P(1) = 1 . 2 = 2 = (1 - 1) 2^{1 + 1} + 2\]
\[\text{ Thus, P(1) is true } . \]
\[\text{ Step 2 } : \]
\[\text{ Let P(m) be true .} \]
\[\text{ Then, } \]
\[1 . 2 + 2 . 2^2 + . . . + m . 2^m = (m - 1) 2^{m + 1} + 2\]
\[\text{ To prove: P(m + 1) is true . } \]
\[\text{ That is, } \]
\[1 . 2 + 2 . 2^2 + . . . + (m + 1) 2^{m + 1} = m . 2^{m + 2} + 2\]
\[\text{ Now, } \]
\[P(m) = 1 . 2 + 2 . 2^2 + . . . + m . 2^m = (m - 1) 2^{m + 1} + 2\]
\[ \Rightarrow 1 . 2 + 2 . 2^2 + . . . + m . 2^m + (m + 1) . 2^{m + 1} = (m - 1) 2^{m + 1} + 2 + (m + 1) . 2^{m + 1} \left[ \text{ Adding} (m + 1) . 2^{m + 1} \text{ to both sides} \right]\]
\[ \Rightarrow P(m + 1) = 2m . 2^{m + 1} + 2 = m . 2^{m + 2} + 2\]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]
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