Question

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Answer 1

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QUESTION:

$\textsf{If S_n denotes the sum of n terms of an AP} \\ \textsf{whose common difference is d and first term is a} \\ \textsf{Find S_n + 2 S_{n - 1} + S_{n - 1} }.$

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ANSWER:

• $\textsf{ S_n + 2 S_{n - 1} + S_{n - 1} } = a(4n - 4) + d(2n^2-2n+5)$

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GIVEN:

• First term = a

• Common difference = d

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TO FIND:

• $\textsf{ S_n + 2 S_{n - 1} + S_{n - 1} }.$

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TERMS USED:

$\boxed{ \bold{\large{S_n = \frac{n}{2} (2a + (n - 1)d)}}}$

$2S_{n-1} = 2\bigg(\dfrac{n - 1}{2} \bigg)( 2a + (n-1-1)d)$

$2S_{n-1} = (n - 1)(2a + (n-2)d)$

$S_{n-2} = \bigg(\dfrac{n - 2}{2} \bigg)2a + (n-2 - 1)d$

$S_{n-2} = \dfrac{n - 2}{2}(2a + (n-3)d)$

NOTE : REFER ATTACHMENT FOR CALCULATION.

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Answer 2

Answer:

Sn = (n/2)[ 2a + ( n -1) d]

then Sn - 2Sn-1 + Sn +  2

⇒ (n/2)[ 2a + ( n -1) d] - 2(n - 1)/2)[ 2a + ( n - 1 -1) d] + (n +2) / 2)[ 2a + ( n + 2 -1) d]

⇒ (1/2)[ 2an + n( n -1) d] + [ 4a(n - 1) + 2(n - 1)( n - 2) d] +[ 2a(n + 2)+ ( n + 1) (n + 2) d]

⇒ (1/2)[ 2a[ n - 2n + 2 + n + 2] + d [ n2 - n - 2n2 + 6n - 4 + n2 + 3n + 2] ]

⇒ (1/2)[ 2a(4) + d(8n - 2) ]

= [ 4a + (4n - 1)d]