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यदि $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots n$ पद $= n(n+1) f(n) - 3n$ है,तो $f(1) =$

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श्रेणी $1 + 6 + 13 + 22 + 33 + \dots$ के $n$ पदों का योग ज्ञात कीजिए।

$\frac{1^3 + 2^3 + 3^3 + 4^3 + \dots + 12^3}{1^2 + 2^2 + 3^2 + 4^2 + \dots + 12^2} = $

यदि $\alpha \in R, n \in N$ और $n+2(n-1)+3(n-2)+\ldots+(n-1)2+n.1 = \alpha n(n+1)(n+2)$ है,तो $\alpha =$

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