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Sum the series up to n terms
(3)(2)2+(5)(3)2+(7)(4)2+…..
Sn=(2n−1)2Sn=\frac{\left(2n-1\right)}{2}Sn=2(2n−1)
Sn=n2(n+1)22Sn=\frac{n^2\left(n+1\right)^2}{2}Sn=2n2(n+1)2
Sn=(n)(3n3+16n2+30n+23)6Sn=\frac{\left(n\right)\left(3n^3+16n^2+30n+23\right)}{6}Sn=6(n)(3n3+16n2+30n+23)
What is the formula for the sum of the first n positive even integers?
2n22n^22n2
n2n^2n2
n
What is the formula for the sum of an arithmetic series with the first term a, common difference d, and n terms?
n2(a+(n−1)d)\frac{n}{2}\left(a+\left(n-1\right)d\right)2n(a+(n−1)d)
(2a+(n−1)d)\left(2a+\left(n-1\right)d\right)(2a+(n−1)d)
n2(2a+(n)d)\frac{n}{2}\left(2a+\left(n\right)d\right)2n(2a+(n)d)
Fill in the blank
∑k=1nk=\sum_{k=1}^nk=∑k=1nk=
n(n+1)2\frac{n\left(n+1\right)}{2}2n(n+1)
n(4n2−1)3\frac{n\left(4n^2-1\right)}{3}3n(4n2−1)
n(n2−1)3\frac{n\left(n^2-1\right)}{3}3n(n2−1)
What is the formula for the sum of the reciprocals of the first n natural numbers?
1n\frac{1}{n}n1
k
∑k=1nk3=\sum_{k=1}^nk^3=∑k=1nk3=
(n(n+1)22)\left(\frac{n\left(n+1\right)^2}{2}\right)(2n(n+1)2)
((n+1)2)2\left(\frac{\left(n+1\right)}{2}\right)^2(2(n+1))2
n(n−1)23\frac{n\left(n-1\right)^2}{3}3n(n−1)2
(2)(1)2+(4)(2)2+(6)(3)2+…..
Sn=(n+1)(2n−1)6Sn=\frac{\left(n+1\right)\left(2n-1\right)}{6}Sn=6(n+1)(2n−1)
∑k=1n1=\sum_{k=1}^n1=∑k=1n1=
1
Sum the series up to n terms. 12+32+52+.....
Sn=n(4n2−1)3Sn=\frac{n\left(4n^2-1\right)}{3}Sn=3n(4n2−1)
Sum the series up to n terms. 121+(12+22)2+(12+22+32)3\frac{1^2}{1}+\frac{\left(1^2+2^2\right)}{2}+\frac{\left(1^2+2^2+3^2\right)}{3}112+2(12+22)+3(12+22+32) +....
sn=(2n−1)2sn=\frac{\left(2n-1\right)}{2}sn=2(2n−1)
sn=(4n2+15n+17)36sn=\frac{\left(4n^2+15n+17\right)}{36}sn=36(4n2+15n+17)
Sn=n(4n2+15n+17)36Sn=\frac{n\left(4n^2+15n+17\right)}{36}Sn=36n(4n2+15n+17)
It is done.