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Question 90 of 95

Q.The value of 2+4+6+…+2n2+4+6+\ldots+2n is:

(a) 2n(2n+1)2\dfrac{2n(2n+1)}{2}
(b) n(n−1)2\dfrac{n(n-1)}{2}
(c) n(n+1)n(n+1)
(d) n(n+1)2\dfrac{n(n+1)}{2}
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2025MCQ· 1mImportance★★★★★
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This is 22 times the sum of the first nn natural numbers.

The series 2+4+6+…+2n2+4+6+\ldots+2n is an arithmetic progression with first term 22, common difference 22, and nn terms, so it equals 2(1+2+…+n)2(1+2+\ldots+n). …

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