Question

# Let Sn=4n∑k=1(−1)k(k+1)2k2 . Then Sn can take value (s)

A
16n2+4n.
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B
16n2+3n.
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C
18n2+4n.
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D
16n2+5n.
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Solution

## The correct option is A 16n2+4n.S=−12−22+32+42....................+(4n−1)2+(4n2)→(1)⇒S=4n2+(4n−1)2−(4n−2)2−(4n−3)2+...........−22−12→(II) ( in reverse order )Adding them -⇒2S=(4n−1)(4n+1)+(4n−3)(4n+1)−(4n−5)(4n+1).......−(3−4n)(4n+1)−1(1−4n)(4n+1)Factoring out (4n+1) all that's left is to complete the following sum ( let's call it A )⇒(4n−1)+(4n−3)−(4n−5)−(4n−7)+.........(3−4n)−(1−4n)=ABut, sum of each 4 consecutive numbers of A is equal to 8, because if we call the first number in each group of for x, then we have⇒x+(x−2)−(x−4)−(x−6)=8As there are 4n numbers in the sum that is equal to A, there are n groups of 4 and ∴A=8nTherefore, we have ⇒(4n+1)A=2S⇒8n(4n+1)=2Sand finally,⇒S=4n(4n+1)=16n2+4nHence, the answer is 16n2+4n.

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