Sum: SN = 1 * 2 * 3 + 2 * 3 * 4 + +n(n+1)(n+2)

Sum: SN = 1 * 2 * 3 + 2 * 3 * 4 + +n(n+1)(n+2)

Sn=1*2*3+2*3*4+…… +n(n+1)(n+2)=(1³+2³+3³+----+n³)+3(1²+2²+3²+---+n²)+2(1+2+3+----+n)=n²(n+1)²/4+n(n+1)(2n+1)/2+n(n+1)=n(n+1)[n(n+1)/4+(2n+1)/2+1)]=n(...

Sum Sn = 1 / 2 + 3 / 2 ^ 2 + 5 / 2 ^ 3 + +2n-1/2^n
Sum Sn = 1 / 2 + 3 / 2 ^ 2 + 5 / 2 ^ 3 + +(2n-1)/2^n.
^It means power
It's better to subtract by dislocation
emergency

Sn=1/2+3/2^2+5/2^3+… +(2n-1)/2^n2Sn=1+3/2+5/2^2+… +(2n-1) / 2 ^ (n-1) when n > = 2, the result of dislocation subtraction is Sn = 1 + 2 [1 / 2 + 1 / 2 ^ 2 + 1 / 2 ^ 3 +... + 1 / 2 ^ (n-1)] - (2n-1) / 2 ^ NSN = 1 + [1 + 1 / 2 + 1 / 2 ^ 2 + 1 / 2 ^ 3 +... + 1 / 2 ^ (n-2)] - (2n-1) / 2 ^ n = 1 + 2-1 / 2 ^ (n-2) - (2

Sum Sn = 1-3 + 5-7 + 9 + What should + [(- 1) ＾ (n-1)] (2n-1) be?

When n is odd, (- 1) ＾ (n-1) = 1, the last digit is positive Sn = 1 + (5-3) + (9-7) + When n is an even number, (- 1) ＾ (n-1) = - 1, the last digit is a negative number Sn = (1-3) + (5-7) + (9-1