Question

Find the sum to n terms of the series 1²+3²+5²+7²+.........​

Answer 1

Let 2k + 1 ne the general term of the sum. If we'll replace k by values from 0 to n, we'll get each odd natural number.

We'll raise to square and we'll expand the binomial:

(2k+1)^2 = 4k^2 + 4k + 1

We'll take the summation:

1^2 + 2^2 +...n^2 = Sum (2k+1)^2 = Sum 4k^2 + Sum 4k + Sum 1

Sum (2k+1)^2 = 4*n*(n+1)*(2n+1)/6 + 4*n(n+1)/2 + n

Sum (2k+1)^2 = [4*n*(n+1)*(2n+1) + 12n*(n+1) + 6n]/6

We'll remove the brackets:

Sum (2k+1)^2 = [4n(n^2 + 3n + 1) + 12n^2 + 12n + 6n]/6

Sum (2k+1)^2 = (4n^3 + 12n^2 + 4n + 12n^2 + 12n + 6n)/6

We'll combine like terms:

Sum (2k+1)^2 = (4n^3 + 24n^2 + 22n)/6

Sum (2k+1)^2 = 2n(2n^2 + 12n + 11)/6

Sum (2k+1)^2 = n(2n^2 + 12n + 11)/3

The requested sum of the squares of the odd natural numbers is: 1^2 + 2^2 + ...n^2 = Sum (2k+1)^2 = n(2n^2 + 12n + 11)/3

Answer 2

Answer:

Step-by-step explanation:

1^2 =1 (1*1=1)

3^2=9. (3*3=9)

5^2=25. (5*5=25)

7^2=49. (7*7=49)

In between you can also have 4,6......

By This method you can find till n

Hope it helps

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