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If $49^{n} + 16 n + λ$ is divisible by 64 for all $n \in N,$ then the least negative integral value of $λ,$ is

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Solution

For $n = 1,$ we have
$49^{n} + 16 n + λ = 49 + 16 + λ = 65 + λ$
$= 64 + (λ + 1),$ which is divisible by 64 if $λ = - 1 .$
For $\begin{array}{l} n = 2 \end{array}$ we have
$49^{n} + 16 n + λ = 49^{2} + 16 \times 2 + λ = 2433 + λ$
$= (64 \times 38) + (λ + 1),$ which is divisible by 64 if $λ = - 1$
Hence, $λ = - 1$ .

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