Observation gives rise to a problem! (Part 2)

Algebra Level 4

$\Large\displaystyle\sum_{r=1}^{10^{1000}} r^2$

What is the value of the sum of digits of the summation above?

Bonus: Generalize for $10^n$ .

Try Part 1 (click here) .

The answer is 6010.

1 solution

Nihar Mahajan
Jun 11, 2015

This is just a pattern and not a proof:

Generalize for $10^n$ by seeking pattern:

When $n=1$ ,sum $=\dfrac{(10)(11)(21)}{6}=385$ , So sum of digits = $3+8+5 = 16$ .

When $n=2$ ,sum $=\dfrac{(100)(101)(201)}{6}=338350$ , So sum of digits = $3(3)+8+5 = 22$ .

When $n=3$ ,sum $=\dfrac{(1000)(1001)(2001)}{6}=333833500$ , So sum of digits = $3(5)+8+5 = 28$ .

$\dots$

When $n=n$ ,sum $=\dfrac{(n)(n+1)(2n+1)}{6}$ , So sum of digits = $3(2n-1)+8+5 = \boxed{6n+10}$ .

Here $n=1000$ which means sum of digits $=6(1000)+10=\Large\boxed{6010}$ .

Moderator note:

This solution is wrong. You have only shown that the pattern appears to work for small $n$ , how do you guarantee that the pattern won't fail at larger $n$ 's?

I also used a similar method as i am not very good in number theory. You can slightly shorten the method by not considering n in $\frac {n (n+1)(2n+1)}{6}$ as it will only increase the no of zeroes at the end.

Satvik Choudhary - 6 years ago

Did any one get 5999?

raman rai - 6 years ago

Observation gives rise to a problem! (Part 2)

Try Part 1 (click here) .

The answer is 6010.

1 solution

Moderator note:

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