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I need the sum of the series $$\sum_{i=1}^t {(t-i+1)^2 a^{2(t-i)} }$$ it doesn't looks like arithmetic or geometric series but mixture of these two.

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Instead of summing it from $i=1$ to $t$, sum it from $i=t$ to $1$.

$$=\sum_{i=1}^ti^2(a^2)^{i-1}=a^{-2}\sum_{i=1}^ti^2(a^2)^i$$

And as per this post, we know that

$$\sum_{i=1}^ti^2x^i={(x (1 + x) - x^{1 + t} ((1 + t)^2 - (-1 + 2 t + 2 t^2) x + t^2 x^2)\over(1 - x)^3}$$

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    Thanks for your help but when I use the above formula for specific case e.g, t=3, a=0.7. Series sum and formula described by you produce different results – 2017-01-04
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    @Ali are you sure you put in the numbers right? Note $x=a^2$. What do you get with those values? – 2017-01-04
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    Dear Sir if I replace $$ 1-x^3$$ by $$ (1-x)^3 $$ in the formula given by you, Then I get the answer equal to the sum of the series. – 2017-01-04
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    @ Simple Art, Yes I used $$x=a^2$$. Now I got the same answer for t=3 and a=0.7. Ans=5.1209 – 2017-01-04
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    @Ali Ah, that is a typo on my part. Sorry. – 2017-01-04
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    Well, glad to help and that it all worked out :D – 2017-01-04
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    Thanks a lot for this favor. – 2017-01-08
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    @Ali Glad to help! :D – 2017-01-08