Question

find the sum of n terms in the series 1.4+3.7+5.10+...........is​

Answer 1

Answer:

$\frac{n}{2}( 4n^2 + 17n + 21)$

Step-by-step explanation:

Given :

To Find the value of :

1  x 4 + 3 x 7 + 5 x 10 + ... (n terms)

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Solution :

1 x 4 + 3 x 7 + 5 x 10 ...

⇒ (2(0) + 1) (3(0) +4) + (2(1) + 1) (3(1) + 4) + ...(2(n-1) + 1) (3(n-1) + 4) + (2n + 1) (3n + 4)

⇒ $\Sigma^n_{x=1} {(2x+1)(3x+4)}$

⇒$\Sigma^n_{x=1} {6x^2+11x+4}$

⇒ $\Sigma^n_{x=1} {6x^2} + \Sigma^n_{x=1} {11x} + \Sigma^n_{x=1} {4}$

⇒ $6 (\frac{n(n+1)(2n+1)}{6}) + 11 (\frac{n(n+1)}{2}) + 4n$

⇒ $n(n+1)(2n+1) +  \frac{11n(n+1)}{2} + 4n$

⇒ $n(n+1)[2n+1 + \frac{11}{2} ] + 4n$

⇒ $n(n+1)[ \frac{4n + 13}{2} ] + 4n$

⇒ $n[(n+1)( \frac{4n + 13}{2} ) + 4]$

⇒ $n[( \frac{4n^2 + 17n + 13}{2} ) + 4]$

⇒ $n[( \frac{4n^2 + 17n + 21}{2} )]$

⇒ $\frac{n(4n^2 + 17n + 21)}{2}$

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∴ 1  x 4 + 3 x 7 + 5 x 10 + ... (n terms) = $\frac{n}{2}( 4n^2 + 17n + 21)$

Answer 2

Answer:

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