Problem that needs hardwork?

The Problem Student got answer by observation.

Hard workers find solution by the formula $\large \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^{n} i^3 = {(\frac{n(n+1)}{2})}^{2}$

But calculators would float.

What irony!

Answer of equation 1 is $333333333338333333333350000000000$
Answer of equation 2 is $25000000500000002500000000000000$
So, their sum is $358333333838333335833350000000000$

---

For equation 1

First i found $\large \sum_{n=1}^{10} n^2$ = 385

$\large \sum_{n=1}^{100} n^2$ = 338350

$\large \sum_{n=1}^{1000} n^2$ = 333833500

$\large \sum_{n=1}^{10000} n^2$ = 333383335000

So, we observe thar in each case one additional 3 is being added to the left and right of 8. Similarly, one additional 0 is being added to the right of 0. When i am finding for 100 there are two 3s on left of 8 and one 3 on right of 8 and one 0 on right of 5. So, for 100000000000, there should be eleven 3s on the left of 8, ten 3s on right of 8 and ten 0s on right of 5.i.e $333333333338333333333350000000000$ .

---

For equation 2

First i found $\sum_{n=1}^{100} n^3$ = 25502500

$\sum_{n=1}^{1000} n^3$ = 250500250000

$\sum_{n=1}^{10000} n^3$ = 2500500025000000

So, we find that in every case an additional 0 is being added to the left and right of the second 5 in each case. Then to the right of the third 5, twice the number of 0s as the njmber of them to the right of the second 5 is present. For, 1000, there is one 0 to the left of the second 5 and two 0s to the right of it. Then there are 2×2 zeroes to right of the third 5 of the number. So, for 100000000, there would be six zeroes to the left of the second 5, seven 0s to the right of them and to the right of the third 5, there would be 7×2=14 zeroes. i.e. $25000000500000002500000000000000$

---

So, the final answer is $333333333338333333333350000000000$ + $25000000500000002500000000000000$ = $358333333838333335833350000000000$