Given sequence is 1,2,4,7,11,16... Find 125th term and also find sum.(good luck)

Clearly, the recursion relation for your sequence is

a

_{n+1}= a_{n}+ n, with a_{1}=1.Since the n-th and (n+1)-th term differ by n, we suspect (from calculus) that the n-th term a

_{n}has as its leading term n²/2. To ensure the leading term is divisible by 2, modify this as n(n-1)/2:a

_{n}=n(n-1)/2 + c for some unknown constant c.After some trial and error, I found c=1, so

a

_{n}=n(n-1)/2+1or a

_{n}=(n²-n+2)/2Therefore, the 125-th term is

a

_{125}=125*124/2+1 = 7751Similarly, you find the partial sum

∑a

_{n}=∑(n²-n+2)/2Again, calculus indicates that the leading term is of the form n³/6. After quite a bit of trial and error, I got

∑a

_{n}=(n³+5n)/6so that

∑

_{n=1}^{125}a_{n}=(125³+5(125))/6 = 325,625.
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