Re: Evaluating the series

[Montgomery-Smith, Stephen]

On 03/08/17 01:32, inor0627 wrote:
> Mphumzi Tshentu wrote
>> Good day Sir/Madam
>>
>> can you please help me with the Octave code that I can use to evaluate the
>> series:
>> 1^2+2^2+3^2+...+100^2
>>
>> Thanks
>> Mphumzi
>>
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>
> Hello Mphumzi,
>
> homework? This should do the trick:
>
> result=1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+13^2+14^2+15^2+16^2+17^2+18^2+19^2+20^2+21^2+22^2+23^2+24^2+25^2+26^2+27^2+28^2+29^2+30^2+31^2+32^2+33^2+34^2+35^2+36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2+45^2+46^2+47^2+48^2+49^2+50^2+51^2+52^2+53^2+54^2+55^2+56^2+57^2+58^2+59^2+60^2+61^2+62^2+63^2+64^2+65^2+66^2+67^2+68^2+69^2+70^2+71^2+72^2+73^2+74^2+75^2+76^2+77^2+78^2+79^2+80^2+81^2+82^2+83^2+84^2+85^2+86^2+87^2+88^2+89^2+90^2+91^2+92^2+93^2+94^2+95^2+96^2+97^2+98^2+99^2+100^2

Very impressive!

Here is my method.

f = @(n)(1/6*n*(n+1)*(2*n+1));
f(100)
ans = 3.3835e+05

Note this is "better" than
f = @(n)(n*(n+1)*(2*n+1)/6);
because the first way changes the answer from an integer to floating point.