Find the sum using the formulas for the sums of powers of integers.
sum_(i=1)^6 (5i-6i^(3))
Which formulas for the sums of powers of integers are needed to find the sum? (Select all that apply.)
1+2+3+4+cdots+n=(n(n+1))/(2)
1^(2)+2^(2)+3^(2)+4^(2)+cdots+n^(2)=(n(n+1)(2n+1))/(6)
1^(3)+2^(3)+3^(3)+4^(3)+cdots+n^(3)=(n^(2)(n+1)^(2))/(4)
1^(4)+2^(4)+3^(4)+4^(4)+cdots+n^(4)=(n(n+1)(2n+1)(3n^(2)+3n-1))/(30)
1^(5)+2^(5)+3^(5)+4^(5)+cdots+n^(5)=(n^(2)(n+1)^(2)(2n^(2)+2n-1))/(12)
So, we have the following.
sum_(i=1)^6 (5i-6i^(3))=sum_(i=1)^6 5i-sum_(i=1)^6 (,)i^(3)
=5((6(6+1))/(2))-6((()^(2)(6+1)^(2))/(4))
=5(21)-6(,)
=
Find the sum using the formulas for the sums of powers of integers.
5i-63
Which formulas for the sums of powers of integers are needed to find the sum? (Select all that apply.) 1+2 =n+1 2 12 + 22 + 32 + 42 + ... + n2 = n(n +1)(2n +1)
13+23 +33+43+...+3=n2n+12
14 + 24 + 34 + 44 + ... + n4 = n(n + 1)(2n + 1)(3n2 + 3n - 1) 30 15 + 25 + 35 + 45 + ... + n5 = n2(n + 1)2(2n2 + 2n - 1) 12
So, we have the following.
(5i - 6i3) = 5i -
66+1
6+12
=521-6