Question

A sequence consists of 20102010 terms. Each term after the first is 11 larger than the previous term. The sum of the 20102010 terms is 53075307. When every second term is added up, starting with the first term and ending with the second last term, the sum is

Answer 1

You're considering a sequence of in which consecutive terms differ by 1, meaning

a(n) = a (n - 1) + 1

so a(n) is an arithmetic sequence. (I'm guessing 20102010 should actually be 2010, and 53075307 should be 5307, so 11 should probably be just 1.)

The sum of the first 2010 terms is 5307, or

$\displaystyle\sum_{n=1}^{2010}a(n)=5307$

Find the value of the first term in the sequence, a(1).

We can write a(n) in terms of a(1) by iterative substitution:

a(n) = a(n - 1) + 1

a(n) = (a(n - 2) + 1) + 1 = a(n - 2) + 2

a(n) = (a(n - 3) + 1) + 2 = a(n - 3) + 3

and so on, down to

a(n) = a(1) + n - 1

So the sum of the first 2010 terms is

$\displaystyle\sum_{n=1}^{2010}a(n)=\sum_{n=1}^{2010}\left(a(1)+n-1\right)=(a(1)-1)\sum_{n=1}^{2010}1+\sum_{n=1}^{2010}n=5307$

Recall that

$\displaystyle\sum_{n=1}^N1=\underbrace{1+1+\cdots+1}_{N\text{ times}}=N$

and

$\displaystyle\sum_{n=1}^Nn=1+2+\cdots+N=\dfrac{N(N+1)}2$

So we have

$\displaystyle\sum_{n=1}^{2010}a(n)=2010(a(1)-1)+\frac{2010\cdot2011}2=5307$

Solve for a(1) :

2010 (a(1) - 1) + 2,021,055 = 5307

2010 (a(1) - 1) = -2,015,748

a(1) - 1 = - 335,958/335

a(1) = - 335,623/335

Now, every second term, starting with a(1), differs by 2, so they form another arithmetic sequence b(n) given by

b(n) = b(n - 1) + 2

or, using the same method as before,

b(n) = b(1) + 2 (n - 1) = a(1) + 2n - 2

The sum of the 1005 terms in this sequence is

$\displaystyle\sum_{n=1}^{1005}b(n)=(a(1)-2)\sum_{n=1}^{1005}1+2\sum_{n=1}^{1005}n$

= (- 335,623/335 - 2)•1005 + 2•1005•1006/2

= 1146