Mathematics in Ancient and Medieval India c7.s2.44

c7.s2.44

(b) Āryabhaṭa I³ (476 A.D.) gave the summation of the sums of n times the 1st term of n natural numbers, (n-1) times the 2nd term of n natural numbers, (n-2) times the 3rd term of natural numbers……etc. Thus if \begin{align*} S_1 &= n.1 \\ S_2 &= (n-1).2 \\ S_3 &= (n-2).3 \\ & \dots\dots\dots\dots\dots\dots \\ & \dots\dots\dots\dots\dots\dots \\ S_{n} &= [ n-(n-1) ].n \end{align*} then according to Āryabhaṭa I, \begin{align*} & S_1 + S*2 + \dots \dots + S*{n} \\ = \sum_{n=1}n S_{n} & \\ = n.1 + (n-1).2 & + (n-2).3 + \dots\dots + \\ & [n-(n-1)].n \\ = 1 + (1+2) & + (1+2+3) + \dots + (1+2+\dots + n) \\ = (n(n+1)(n+2))/6 & \end{align*}