Question

The number 65 is circled. (65.) Find the sum: $\cos \theta + 2^2 \cos 2\theta + 3^2 \cos 3\theta + ... + n^2 \cos n\theta$.

Objective Questions

Answer 1

$\sum_{k=1}^{n}k^2\cos k\theta =\frac{\sin\frac{n\theta}{2}}{4\sin^3\frac{\theta}{2}}\Biggl[ 2\sin\frac{\theta}{2}\cos\frac{(2n+1)\theta}{2} -(2n+1)\cos\frac{\theta}{2}\sin\frac{(2n+1)\theta}{2}\Biggr] +\frac{n(n+1)}{2}\frac{\sin n\theta}{\sin^2\frac{\theta}{2}}.$

Answer 2

We will show that the answer can be written in “closed‐form”. One acceptable answer was

$$\sum_{k=1}^{n}k^2\cos k\theta =\frac{\sin\frac{n\theta}{2}}{4\sin^3\frac{\theta}{2}}\Biggl[ 2\sin\frac{\theta}{2}\cos\frac{(2n+1)\theta}{2} -(2n+1)\cos\frac{\theta}{2}\sin\frac{(2n+1)\theta}{2}\Biggr] +\frac{n(n+1)}{2}\frac{\sin n\theta}{\sin^2\frac{\theta}{2}}\,.$$

Below is a brief explanation of one way to arrive at such an answer.

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Minimal Explanation

1. Write the Sum in Complex Form:
    Express the cosine as the real part:
     $\cos k\theta =\Re\bigl(e^{ik\theta}\bigr).$  Thus   $\sum_{k=1}^{n}k^2\cos k\theta=\Re\Bigl(\sum_{k=1}^{n} k^2e^{ik\theta}\Bigr).$

2. Use a Generating Function:
    Recall that for $|x|\ne1$ one may show   $\sum_{k=1}^{n}k^2x^k=\frac{x(1+x)}{(1-x)^3}  -\; x^{n+1}\Biggl[\frac{n^2+n}{1-x}+\frac{2n+1}{(1-x)^2}+\frac{2}{(1-x)^3}\Biggr].$  Now substitute $x=e^{i\theta}$ and take the real part.

3. Simplify to Trigonometric Functions:
    After some algebra (using the standard identities   $\displaystyle 1-e^{i\theta}=2ie^{i\theta/2}\sin\frac{\theta}{2}$  and its consequences) one obtains the answer in terms of sine and cosine functions.

Since there is usually more than one algebraically equivalent form, the answer given above is acceptable.