Don't understand what they want me to do...part d?

Question

an=n(n+1)
Consider the Series Sn = a1 + a2 +a3 + ... + an where ak is defineds as above.
a) determine several values of Sk, including S1, S2, S3..., S6 and note observations
b) Formulate a conjexture for a general expression for Sn
c) Prove you conjecture by induction.
d) Using the above result, calculate 1^2 + 2^2 + 3^2 + 4^2 + .... + n^2

Answer 1

If the problem is really only part d, that's simple enough. Having proved your result for Sn in c, use that to calculate the given sum.

Sn = a1 + a2 + a3 + ... + an
= 1(1+1) + 2(2+1) + 3(3+1) + ... + n(n+1)
= 1(1) + 2(2) + 3(3) + ... + n(n) + 1(1) + 2(1) + 3(1) + ... + n(1)
= (1^2 + 2^2 + ... + n^2) + (1 + 2 + ... + n)
So 1^2 + ... + n^2 = Sn - (1 + 2 + ... + n), and you should know or be easily able to prove that (1 + 2 + ... + n) = n(n+1)/2. Since you've proven an expression for Sn, you can put that in to give you an expression for 1^2 + 2^2 + ... + n^2.

If you want help with parts (a)-(c), that will take longer. ;-)

Answer 2

here comes: s1=2; s2=8; s3=20; s4=40; s5=70; s6=112; s[k+1] = s[k]+k(k+1); I can’t make it out, but let’s consider
C(n+1) = 1^3 + 2^3 +3^3 +4^3 + = Sum{for k=1 until n+1} of k^3 = 1 + Sum{for k=1 until n} of (1+k)^3 = 1 + Sum{for k= until n} of (1+3k+3k^2+k^3) = 1 + 3(Sum of k) +3(Sum of k^2) + C(n) = 1 + n + 3B(n) + 3D(n) + C(n)
On the other hand: C(n+1)=C(n) + (n+1)^3; thus 1+n + 3B(n) + 3D(n) = (n+1)^3, where
B(n) = 1+2+3+…+n = Sum{for k=1 until n} of k = n(n+1)/2; just add pairs (1st+last), (2nd+(n-1)) and so on!, number of pairs being n/2;
Hence D(n) = 1^2+2^2 +3^2+… +n^3 = ((n+1)^3 – (n+1) – (3/2)(n+1)n)/3 = n(n+1)(2n+1)/6;

Our task is: S(n) = Sum{for k=1 until n} of k(k+1) = (Sum of k^2) + (Sum of k) = D(n) + B(n) = n(n+1)(1/2+(2n+1)/6) = 2n(n+1)(n+2)/6;

it could help i hope

Answer 3

Series, plug in values
a(0)=0(0+1)=0
a(1)=1(1+1)=2