Make the correct alternative in the following question:

We have,

$\mathrm{P}(n): 49^{n}+16^{n}+\lambda$ is divisible by 64 for all $n \in \mathbf{N}$.

For $n=1$,

$\mathrm{P}(1)=49^{1}+16^{1}+\lambda=65+\lambda$

As, the nearest value of $\mathrm{P}(1)$ which is divisible by 64 is 64 itself.

$\Rightarrow 65+\lambda=64$

$\Rightarrow \lambda=64-65$

$\therefore \lambda=-1$

Hence, the correct alternative is option (c).