The sum of series 1²/2 + 1²+2²/6 + 1²+2²+3²/12 + 1²+2²+3²+4²/20 + upto 30 terms is
Answers
Answer: The sum is:
∑k=1nk2=n6(n+1)(2n+1) .
The formula is derived by the following procedure.
Preparation: Let us derive the formula for ∑k as a preparation. In order to obtain the formula, let us consider the following sum:
S=12+22+32+⋯+n2 .
If we subtract 1 from the series and then add (n+1)2 to the series, then obtaining
S+(n+1)2−1=22+32+⋯+n2+(n+1)2 .
The difference between the second and the first series is written in the form:
n2+2n=∑k=1n[(k+1)2−k2]=2∑k=1nk+2∑k=1n1=2∑k=1nk+n .
Finally, we can identify the sum as
∑k=1nk=n2(n+1) .
Derivation of sum of second order: ∑k2 may be obtained by the similar procedure. First, let us consider the series of the third order:
S=13+23+33+⋯+n3 .
Subsequently, we subtract 1 form the series, which then is added by (n+1)3 ; we then write the result as
S+(n+1)3−1=23+33+⋯+n3+(n+1)3.
The difference between the second and the first series is written in the form:
n3+3n2+3n=∑k=1n[(k+1)3−k3]=3∑k=1nk2+3∑k=1nk+n .
With the help of the preparation we made in advance, we can write
3∑k=1nk2=n3+3n2+3n−n2(n+1)−n=n3+32n2+12n .
Finally, we successfully obtain the formula of the form:
∑k=1nk2=n6(n+1)(2n+1) .