arithmatic progression

Question

(1+3)+(5+7+9+11)+(13+15+17+19+21+23)....is an A.P series. find the sum upto nth term.

Answer 1

The parenthesis are irrelevant.

 

For an arithmetic series:

 

S_n = (n/2)(2a₁ + (n - 1)d)

 

S_n = (n/2)(2 + (n - 1)2)

S_n = (n/2)(2 + 2n - 2)

S_n = (n/2)(2n)

S_n = n²

Answer 2

I'm assuming (1+3) is the first term, (5+7+9+11) is the second term, and so on... correct ?

Each term contains an even number of odd numbers.

term 1 contains 2 odd numbers

term 2 contains 4 odd numbers

term 3 contains 6 odd numbers

term 4 contains 8 odd numbers

and so on...

other facts...

the sum of the first 2 odd numbers is 2^2=4

the sum of the first 3 odd numbers is 3^2=9

the sum of the first 4 odd numbers is 4^2=16

the sum of the first "n" odd numbers is n^2

do you see the pattern ?

getting back to the terms...

term 1 contains 2*1=2 odd numbers

term 2 contains 2*2=4 odd numbers

term 3 contains 2*3=6 odd numbers

term 4 contains 2*4=8 odd numbers

and so on...

term 1+ term 2 contains 6 odd numbers altogether

term 1+ term 2+ term 3 contains 12 odd numbers altogether

terms 1+2+3+4 contain 20 odd numbers altogether

terms 1+2+3+4+5 contains 30 odd numbers altogether (1 has 2, 2 has 4, 3 has 6, 4 has 8, and 5 has 10 for a total of 30 odd numbers)

do you see the pattern ?

terms 1+2 contain 2*3=6 odd numbers

terms 1+2+3 contain 3*4=12 odd numbers

terms 1+2+3+4 contain 4*5=20 odd numbers

terms 1+2+3+4+5+...+n contain n*(n+1) odd numbers or n²+n odd numbers

remember that the sum of the first n odd numbers is n² ?

therefore, the sum of the first (n²+n) odd numbers is (n²+n)²

(n²+n)²=n⁴+2n³+n² which is the answer to your question

for example, suppose n is 3

3⁴+2*3³+3²=81+54+9=144

the first three terms contain the following odd numbers...

(1+3)+(5+7+9+11)+(13+15+17+19+21+23)=12²=144, or...

4+32+108=144