Question

यदि किसी समान्तर श्रेणी के प्रथम n पदों का योग cn? है, तब इन n पदों के वर्गों का योग निम्न है [IIT-JEE - 2009,Paper-2, (3,-1),80] (A) (4n' - 1)c2 n(4n +1) (B) n(4n2 - 1) n(4n? +1) 6 (C) 3 (D) 3 6 answer this question ❓​

Answer 1

If the sum of first n terms of an AP is cn2, then the sum of squares of these n terms is

(1) n(4n2-1)c2/6

(2) n(4n2+1)c2/3

(3) n(4n2-1)c2/3

(4) none of these

Solution:

Given Sn = cn2

Sn-1 = c(n-1)2

= cn2+c-2cn

Tn = Sn-Sn-1

Tn = 2cn – c

Tn2 = (2cn-c)2

= 4c2n2+c2-4c2n

Sum of squares = ∑Tn2

= c2[4 ∑n2+∑1 -4∑n]

= [4c2(n(n+1)(2n+1)/6] + nc2 – (2c2n(n+1)

= nc2(4n2+6n+2+3-6n-6)/3

= nc2(4n2-1)/3

= n(4n2-1)c2/3

Hence option (3) is the answer

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