Question

The sum of first n terms of the series 1^2+2*2^2+3^2+2*4^2+5^2+2*6^2+.....isn(n+1)^2, when n is even, when n is odd the sum is

Answer 1

Heya User,

--> 1² + 2.2² + 3² + 2.4² + 5² + ... 
= [ 1² + 3² + 5² + 7² + ... ] + 2[ 2² + 4² + 6² + ... ]

Now, if n--> an odd -->
=> [ 1² + 3² + ... + n² ] + 2 [ 2² + 4² + 6² + ... + (n-1)² ]
= [ 1² + 2² + ... + n² ] + [ 2² + 4² + 6² + ... + (n-1)² ]
= [tex] \frac{n(n+1)(2n+1)}{6} + 4 [ \frac{ \frac{(n-1)}{2} ( \frac{(n-1)}{2}+1)(n-1+1) }{6} ]\\\\ = \frac{n(n+1)(2n+1)}{6} + \frac{ n(n-1) (n+1) }{6} \\\\ = \frac{n(n+1)[2n+1+n-1]}{6} \\\\ = 3n^2 (n+1)/6 = n^2(n+1)/2[/tex]

Now, if n is even -->
=> [ 1² + 3² + ... + (n-1)² ] + 2 [ 2² + ... + n² ]
= [ 1² + 2² + ... + n² ] + 4 [ 1² + 2² + ... + (n/2)² ]
= [tex] \frac{n(n+1)(2n+1)}{6} + 4[ \frac{\frac{n}{2} [\frac{n}{2} +1][n+1]}{6} ]\\\\ = \frac{n(n+1)(2n+1)}{6} + \frac{ n [n+2][n+1]}{6} \\\\ = \frac{n(n+1)[2n+1+n+2]}{6} \\\\ = \frac{n(n+1)(3n+3)}{6} = \frac{n (n+1)^{2} }{2} [/tex]

So, There's your answer....

Answer 2

Answer:

Step-by-step explanation: