1. S_n = a(a+d)+(a+d)(a+2d)+......+{a+(n−1)d}{a+(n)d} 2. S_n = a(a+d)(a+2d)+(a+d)(a+2d)(a+3d)+......+{a+(n−1)d}{a+(n)d}{a+(n+1)d}


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Question Number 202774 by BaliramKumar last updated on 03/Jan/24
$1. S_n = a(a+d)+(a+d)(a+2d)+......+{a+(n−1)d}{a+(n)d} 2. S_n = a(a+d)(a+2d)+(a+d)(a+2d)(a+3d)+......+{a+(n−1)d}{a+(n)d}{a+(n+1)d}$
$1 . S_{n} = a (a + d) + (a + d) (a + 2 d) + . . . . . . + {a + (n - 1) d} {a + (n) d}$ $2 . S_{n} = a (a + d) (a + 2 d) + (a + d) (a + 2 d) (a + 3 d) + . . . . . . + {a + (n - 1) d} {a + (n) d} {a + (n + 1) d}$
	Answered by Rasheed.Sindhi last updated on 03/Jan/24
	$Σ_(k=1) ^(k=n) {a+kd)}{(a+(k+1)d} Σ_(k=1) ^(k=n) {a^2 +(2k+1)ad+k(k+1)d^2 } Σ_(k=1) ^n a^2 +Σ_(k=1) (2k+1)ad+Σ_(k=1) ^(k=n) k(k+1)d^2 na^2 +Σ_(k=1) (2k+1)ad+Σ_(k=1) ^(k=n) k(k+1)d^2 na^2 +adΣ_(k=1) ^(k=n) (2k+1)+d^2 Σ_(k=1) ^(k=n) k^2 +k) na^2 +ad{2Σ_(k=1) ^(k=n) k+Σ_(k=1) ^(k=n) 1}+d^2 {Σ_(k=1) ^(k=n) k^2 +Σ_(k=1) ^(k=n) k) na^2 +ad{n+2(((n(n+1))/2))}+d^2 {((n(n+1)(2n+1))/6)+((n(n+1))/2)} na^2 +ad{n^2 +2n}+d^2 {((n(n+1)(2n+1)+3n(n+1))/6)} na^2 +n(n+2)ad+(((n(n+1)(2n+1+3))/6))d^2 na^2 +n(n+2)ad+(((n(n+1)(n+2))/3))d^2$
	$\underset{k = 1}{\sum^{k = n}} {a + k d)} {(a + (k + 1) d}$ $\underset{k = 1}{\sum^{k = n}} {a^{2} + (2 k + 1) a d + k (k + 1) d^{2}}$ $\underset{k = 1}{\sum^{n}} a^{2} + \underset{k = 1}{Σ} (2 k + 1) a d + \underset{k = 1}{\overset{k = n}{Σ}} k (k + 1) d^{2}$ ${n a}^{2} + \underset{k = 1}{Σ} (2 k + 1) a d + \underset{k = 1}{\overset{k = n}{Σ}} k (k + 1) d^{2}$ ${n a}^{2} + a d \underset{k = 1}{\overset{k = n}{Σ}} (2 k + 1) + d^{2} \underset{k = 1}{\overset{k = n}{Σ}} k^{2} + k)$ ${n a}^{2} + a d {2 \underset{k = 1}{\overset{k = n}{Σ}} k + \underset{k = 1}{\overset{k = n}{Σ}} 1} + d^{2} {\underset{k = 1}{\overset{k = n}{Σ}} k^{2} + \underset{k = 1}{\overset{k = n}{Σ}} k)$ ${n a}^{2} + a d {n + 2 (\frac{n (n + 1)}{2})} + d^{2} {\frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}}$ ${n a}^{2} + a d {n^{2} + 2 n} + d^{2} {\frac{n (n + 1) (2 n + 1) + 3 n (n + 1)}{6}}$ ${n a}^{2} + n (n + 2) a d + (\frac{n (n + 1) (2 n + 1 + 3)}{6}) d^{2}$ ${n a}^{2} + n (n + 2) a d + (\frac{n (n + 1) (n + 2)}{3}) d^{2}$
	Commented by BaliramKumar last updated on 03/Jan/24

	$Thanks Sir$
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