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.
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 triangulat number is n*(n+1)/2 The 100th is 100*101/2 = 5050
The nth even number is 2n...
The nth triangulat number is n*(n+1)/2. So, the 61st is 61*62/2 = 1891