Question

1.2 + 2.2² + 3.2³ + ... + n.2ⁿ = (n − 1) 2ⁿ⁺¹+2

 

Answer 1

Let P(n) be the given statement.
Now,

\[P(n) = 1 . 2 + 2 . 2^2 + 3 . 2^3 + . . . + n . 2^n = (n - 1) 2^{n + 1} + 2\]

\[\text{ Step 1} : \]

\[P(1) = 1 . 2 = 2 = (1 - 1) 2^{1 + 1} + 2\]

\[\text{ Thus, P(1) is true }  . \]

\[\text{ Step 2 } : \]

\[\text{ Let P(m) be true .} \]

\[\text{ Then, } \]

\[1 . 2 + 2 . 2^2 + . . . + m . 2^m = (m - 1) 2^{m + 1} + 2\]

\[\text{ To prove: P(m + 1) is true . } \]

\[\text{ That is, } \]

\[1 . 2 + 2 . 2^2 + . . . + (m + 1) 2^{m + 1} = m . 2^{m + 2} + 2\]

\[\text{ Now, } \]

\[P(m) = 1 . 2 + 2 . 2^2 + . . . + m . 2^m = (m - 1) 2^{m + 1} + 2\]

\[ \Rightarrow 1 . 2 + 2 . 2^2 + . . . + m . 2^m + (m + 1) . 2^{m + 1} = (m - 1) 2^{m + 1} + 2 + (m + 1) . 2^{m + 1} \left[ \text{ Adding}  (m + 1) . 2^{m + 1} \text{ to both sides}  \right]\]

\[ \Rightarrow P(m + 1) = 2m . 2^{m + 1} + 2 = m . 2^{m + 2} + 2\]

\[\text{ Thus, P(m + 1) is true } . \]

\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]