Question

please tell me the solution for question no. 9​

Answer 1

Answer:

Step-by-step explanation:

When n is odd, let n = 2m +1

∴ The req. sum

= 12 + 2.22 + 32 + 2.42 + ….. + 2(2m) 2 + (2m + 1) 2

= Σ (2m + 1) 2 + 4[12 + 22 + 32 + …. +m2 ]

=(2m + 1) (2m + 2) (4m + 2 + 1)/6 + 4m (m + 1) (2m + 1)/6

= (2m + 1) (m + 1)/6 [2 (4m + 3) + 4m ]

= (2m + 1) (2m + 2)(6m + 3)/6 = ((2m + 1)2 (2m + 2)/2

= n2(n + 1)/2 [ ∵ 2m + 1 =n]

ALTERNATE SOLUTION:

∵ n is odd, last term = n2

∴ req. sum

= [12 + 2.22 + 32 + 2.42 +….. + 2(n-1) 2 ] + n2

= (n – 1)n2 /2 +n2 [ Using sum for( n – 1 ) to be even = n2 (n + 1)/2