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One common example of the "generalize the problem" strategy is found in parameter differentiation.

$\int^{\infty}_{0}\frac{\sin^2(x)}{x^2}dx$

We then create a more general case of $I(a)=\int^{\infty}_{0}\frac{\sin^2(ax)}{x^2}dx$ where $a>0$.

$I'(a)=\int^{\infty}_{0}\frac{2\sin(ax)\cdot\cos(ax)\cdot x}{x^2}dx$

$=\int^{\infty}_{0}\frac{\sin(2ax)}{x}dx$

So $I(a)=1/2\pi a$ and $I(1)=1/2 \pi$

Another similar cute problem is evaluate $\sum_{k=1}^n\frac{k^2}{2^k}$, we instead evaluate $S(x)=\sum_{k=1}^nk^2x^k.$

Edit: Just to finish the second example:

We know that $S(x)=\sum_{k=1}^nx^k=\frac{1-x^{n+1}}{1-x}.$ We take the derivative of this, then multiply that whole mess by x, and take the derivative again, and multiply by x again.

We get $S(x)=\sum_{k=1}^nk^2x^k= \frac{x(1+x)-x^{n+2}-x^{n+1}(nx-n-1)^2}{(1-x)^3}$, so $S(\frac{1}{2})=6-(\frac{n^2+4n+6}{2^n})$.

One common example of the "generalize the problem" strategy is found in parameter differentiation.

$\int^{\infty}_{0}\frac{\sin^2(x)}{x^2}dx$

We then create a more general case of $I(a)=\int^{\infty}_{0}\frac{\sin^2(ax)}{x^2}dx$ where $a>0$.

$I'(a)=\int^{\infty}_{0}\frac{2\sin(ax)\cdot\cos(ax)\cdot x}{x^2}dx$

$=\int^{\infty}_{0}\frac{\sin(2ax)}{x}dx$

So $I(a)=1/2\pi a$ and $I(1)=1/2 \pi$

Another similar cute problem is evaluate $\sum_{k=1}^n\frac{k^2}{2^k}$, we instead evaluate $S(x)=\sum_{k=1}^nk^2x^k.$