पाटीगणित (श्रीधराचार्य - प्राचीन संस्कृत टीका एवं प्रो. कृपाशंकर शुक्ल सविस्तार व्याख्या)

SERIES OF SQUARES, CUBES AND SUCCESSIVE SUMS 85
common difference and number of terms, as before, add to
it the sum of the same (arithmetic) series, and (then) reduce
that (resulting sum) to half : the result (thus obtained) is the
sum of sums of the natural series (beginning with 1 and ending
with the terms of the given arithmetic series).¹
That is to say,
(1+2+3+...to a terms) + (1+2+3+... to a+d terms)+
( +2+3+... to a+2d terms)+... to n terms.
r=n {a+ (r - 1)d} {a+(r - 1)d+1}
≡ ∑ ─────────────────────────────
r=1 2
r=n r=n
= ½ [ ∑ (a + r - 1d)² + ∑ (a + r - 1d) ] .
r=1 r=1
Ex. 120. O the best of mathematicians, say the sum of
the sums of series of natural numbers (each beginning with 1),
whose number of terms are the first six terms of the arithmetic
series with 3 as the first term and 5 as the common difference.
Rule for finding the sum of cubes of the terms of an arithme-
tic series :
107. To the square of the sum of the (given arithmetic)
series, as multiplied by the common difference, add the
product of the first term and the sum of the series, as multi-
plied by the first term minus the common difference : the
result is the sum of cubes of the terms of the (arithmetic)
series with given first term and common difference.²
That is,
a³ +(a+d)³ +(a+2d)³ +...+{a+(n - 1)d }³
r=n
= ∑ {a+(r - 1)d}³
r=1
= S² × d + aS × (a - d),
r=n
where S = ∑ {a+(r - 1)d} .
r=1
Ex. 121. Say after adding together the cubes of the
four terms, which begin with 5 and increase successively by 2.
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¹ For other rules, see GSS, vi. 305-305½ and GK, I, p. 117, lines 11-16.
² Cf. GSS, vi. 303; GK, I, p. 121, lines 2-3, and p. 122, lines 1-2.