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Question Number 202774 by BaliramKumar last updated on 03/Jan/24

1 . S_{n} = a (a + d) + (a + d) (a + 2 d) + \dots \dots + {a + (n - 1) d} {a + (n) d}

2 . S_{n} = a (a + d) (a + 2 d) + (a + d) (a + 2 d) (a + 3 d) + \dots \dots + {a + (n - 1) d} {a + (n) d} {a + (n + 1) d}

Answered by Rasheed.Sindhi last updated on 03/Jan/24

\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