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The question is to find out the value of $$ \cos a+n\cos2a+\frac{n(n-1)\cos3a}{2!}+...$$ upto n terms.

I tried to use the series expansion of $(1+x)^n$ but couldn't manipulate it to bring it in the form required.Any hints to move ahead shall be highly appreciated.Thanks.

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    Adding $\cos((n+1)a)$ to this sum, one gets the real part of $$\sum_{k=0}^n{n\choose k}e^{i(k+1)a}=e^{ia}(1+e^{ia})^n=2^ne^{ia(n+2)/2}\cos(na/2)$$ Can you finish? – 2017-02-06
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    @Did Thanks for your comment.But I didn't get how you simplified $e^{ia}(1+e^{ia})^n$. – 2017-02-06
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    @Did I am afraid.I didn't get.Is this an identity?Please provide an intuition behind this. – 2017-02-06
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    @navinstudent check this https://en.wikipedia.org/wiki/Euler's_formula#Relationship_to_trigonometry – 2017-02-06
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    Yes, this is an identity, note that $$1+e^{ix}=2e^{ix/2}\frac{e^{ix/2}+e^{-ix/2}}2=2e^{ix/2}\cos(x/2)$$ – 2017-02-06
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    To the OP; Are you asking how the summation turned into $e^{ia}(1+e^{ia})^n$ ? It's just by the Binomial Theorem. – 2017-02-06
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    @user254665 yes I was having doubt on how on how to find out the real part of $(1+e^{ix}^n$ it is now cleared.thanks. – 2017-02-07

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$\displaystyle\sum_{r=1}^n\binom nr\cos(r+1)\alpha=$ real part of $\displaystyle\sum_{r=1}^n\binom nr e^{i(r+1)\alpha}$

$\displaystyle\sum_{r=1}^n\binom nre^{i(r+1)\alpha}=e^{i\alpha}\sum_{r=1}^n\binom nr (e^{i\alpha})^r$

$\sum_{r=1}^n\binom nr (e^{i\alpha})^r=-1+(1+e^{i\alpha})^n$

Now, $1+e^{i\alpha}=1+\cos\alpha+i\sin\alpha=2\cos\dfrac\alpha2\left(\cos\dfrac\alpha2+i\sin\dfrac\alpha2\right)$

Now use de Moivre's formula and Euler's formula

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    There is a typo in the answer, the question is about $$\displaystyle\sum_{r=0}^n\binom nr\cos (r+1)\alpha$$ – 2017-02-06
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    Thanks for your answer.A little typo.I think we have to find out $$\sum_{r=1}^{n} \binom {n}{r-1} cos ra$$ and not $$\sum_{r=1}^{n} \binom {n}{r} cos ra$$ – 2017-02-06
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    @navinstudent, Thanks for your observation – 2017-02-06
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    @rtybase, Thanks for your observation – 2017-02-06