solve reccurrence doubt

Question

solve   : Given T(1) =1 

$T(n) = 8T(\frac{n}{2}) + n^2$  

$T(\frac{n}{2}) = 8T(\frac{n}{2^2}) + (\frac{n}{2})^2$

$T(\frac{n}{2^2}) = 8T(\frac{n}{2^3}) + (\frac{n}{2^2})^2$

$T(\frac{n}{2}) = 8^2T(\frac{n}{2^3}) + 8(\frac{n}{2^2})^2 +(\frac{n}{2})^2$

Now, T(n) =  $8^3T(\frac{n}{2^3}) + 8^2(\frac{n}{2^2})^2 + 8(\frac{n}{2})^2 +(\frac{n}{2^0})^2$

now i always stuck after this  help plz , now when we check that which series is forming we have to take  $8^3T(\frac{n}{2^3})$  

seperate from $8^2(\frac{n}{2^2})^2 + 8(\frac{n}{2})^2 +(\frac{n}{2^0})^2$  and then  check  whic series is formed by $8^2(\frac{n}{2^2})^2 + 8(\frac{n}{2})^2 +(\frac{n}{2^0})^2$ 

or we have to take  whole $8^3T(\frac{n}{2^3}) + 8^2(\frac{n}{2^2})^2 + 8(\frac{n}{2})^2 +(\frac{n}{2^0})^2$

Comments

@sumit, now make generalize case

such as $T(n) = 8^kT(\frac{n}{2^k}) + 8^{k-1}(\frac{n^2}{2^{k+1}}) + 8^{k-2}(\frac{n^2}{2^{k-1}})....$

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@ Ashwin Kulkarni how you wrote $2^kT(\frac{n}{2^K})$ , 

if i write $8^3T(\frac{n}{2^3}) = 2^9T(\frac{n}{2^3})$ ( not same) 

i think it should be  , 

$8^KT(\frac{n}{2^k}) +8^{k-1}(\frac{n}{2^{k-1}})^2 + 8^{k-2}(\frac{n}{2^{k-2}})^2+8^{k-k}(\frac{n}{2^{k-k}})^2$

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Sorry that was 8.