If the general term formula of sequence an is that an is equal to the square of two n minus one part of N, then the first five terms are

If the general term formula of sequence an is that an is equal to the square of two n minus one part of N, then the first five terms are

an=n^2- 1/n
a1= 0
a2 = 4- 1/2 = 7/2
a3=9 -1/3 = 26/3
a4=16-1/4 = 63/4
a5=25-1/5 = 24/5

Given the sequence an = n ^ (an equals the square of n), find the sum of sequence Sn =?

(n+1)³-n³=3n²+3n+1
n³-(n-1)³=3(n-1)²+3(n-1)+1
……
2³-1³=3×1²+3×1+1
Add, left middle positive and negative offset
(n+1)³+1=3(1²+2²+…… +n²)+3(1+2+…… +n)+n
1+2+…… +n=n(n+1)/2
Put it into order
1²+2²+…… +n²=n(n+1)(2n+1)/6

LIM (x to 0) SiNx (COS 1 / x) =? Please be as detailed as possible

LIM (x tends to 0) SiNx (cos1 / x) = LIM (x tends to 0) (SiNx / x) * [x (cos1 / x)] = 1 * 0 = 0
LIM (x tends to 0) (SiNx / x) = 1 is an important limit,
LIM (x tends to 0) [x (COS 1 / x)] = 0 uses the property that the product of infinitesimal and bounded variable is still infinitesimal