Question

Prove that 1 + 2 + 2² + ... + 2ⁿ = 2ⁿ⁺¹ - 1 for all n  \[\in\] N .

 

Answer 1

\[Let p\left( n \right): 1 + 2 + 2^2 + . . . + 2^n = 2^{n + 1} - 1 \forall  n \in N\]
\[\text{ Step I: For } n = 1, \]
\[LHS = 1 + 2^1 = 3\]
\[RHS = 2^{1 + 1} - 1 = 2^2 - 1 = 4 - 1 = 3\]
\[As, LHS = RHS\]
\[\text{ So, it is true for n } = 1 . \]
\[\text{ Step II: For n } = k, \]
\[\text{ Let } p\left( k \right): 1 + 2 + 2^2 + . . . + 2^k = 2^{k + 1} - 1\text{  be true }  \forall   k \in N\]
\[\text{ Step III: For } n = k + 1, \]
\[LHS = 1 + 2 + 2^2 + . . . + 2^k + 2^{k + 1} \]
\[ = 2^{k + 1} - 1 + 2^{k + 1} \left(\text{  Using step } II \right)\]
\[ = 2 \times 2^{k + 1} - 1\]
\[ = 2^{k + 1 + 1} - 1\]
\[ = 2^{k + 2} - 1\]
\[RHS = 2^\left( k + 1 \right) + 1 - 1 = 2^{k + 2} - 1\]
\[As, LHS = RHS\]
\[\text{ So, it is also true for n } = k + 1 .\]

Hence, 1 + 2 + 2² + ... + 2ⁿ = 2ⁿ⁺¹ - 1 for all n \[\in\] N .