The sum of the first n terms of the series 12 + 2.22 +32 + 2.4 + 52 + 2.62 +isn(n+1)2 / 2, when n is even. When n is odd, the sum is ………….

Hello Student, Please find the answer to your question Ans. When n is odd, let n = 2m +1 &there4; 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 &there4; 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 ThanksDeepak PatraaskIITians Faculty

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The sum of the first n terms of the series 1² + 2.2² +3²+ 2.4 + 5²+ 2.6²+is
n(n+1)² / 2, when n is even. When n is odd, the sum is ………….

Shane Macguire

12 Years agoGrade upto college level

3 Answers

Deepak Patra

12 Years ago

Hello Student,

Please find the answer to your question

Ans. 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

Thanks
Deepak Patra
askIITians Faculty

Ritik

9 Years ago

Please explain these stepes. :- 12 + 2.22 + 32 + 2.42 + ….. + 2(2m) 2 + (2m + 1) 2= Σ (2m + 1) 2 + 4[12 + 22 + 32 + …. +m2 ]

ankit singh

5 Years ago

When n is odd, it means the sum will be S(n - 1) - the last odd term ie n^2. We can get the result by substituting k = n - 1 in the S(k) expression, and subtracting n ^ 2 from the value, and simplifying it. Hope this answers your question.