Three ways..
Multiply n by itself.
Calculate Sum[2i+1,{i,0,n-1}]
Calculate Sum[n,{i,1,n}]
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One relationship is that the sum of the nth and the previous triangular numbers is equal to the nth square number.That isT(n-1) + T(n) = S(n)where T(n) is the nth triangular number and S(n) is the nth square number.
The nth triangulat number is n*(n+1)/2 The 100th is 100*101/2 = 5050
If your calculator has an exponentiation function, simply raise the number to the power of .5 Remember this trick: the nth root of X = X ^ (1/n)
The nth even number is 2n...
The nth triangulat number is n*(n+1)/2. So, the 61st is 61*62/2 = 1891