Question

#Quality Question

@Sequence And Series

Q]]The sum of the series
$1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + ...$
upto 11th term is​

Answer 1

The sum of the series. 1 + 2 x 3 + 3 x 5 + 4 x 7 +... or 1 + 2*3 + 3*5 + 4*7 +... upto 11th term

Answer:

946

Step-by-step explanation:

Notice the pre-factor is defined by nth term = n and the post factor is just 1 less than twice of it(2n - 1). Thus, nth term of seq is n(2n - 1).

nth term = 2n² - n

Therefore, we need to find

= sum of (2n² - n)

= Sum of seq. 2n² - sum of seq n

= 2 sum of n²(s) - sum of n(s)

= 2 n(n + 1)(2n + 1)/6 - n(n + 1)/2

For n = 11:

= 2(11)(11 + 1)(22 + 1)/6 - 11(11 + 1)/2

= 1012 - 66

= 946

Answer 2

Answer:

• The sum of the given series upto 11th terms is 946

Step-by-step explanation:

Given :-

• ${ \underline{ \boxed{ \tt{1 + 2 \times 3 + 3 \times 5 + 4 \times 7...}}}}$

To Find :-

• The sum of the following series upto the 11th term

Solution :-

✰ We know that,

${\pink{\pmb{\bigstar{\underline{\boxed {\sf \: t _{n} = 2(2n - 1) }}}}}}$

✰According to the question,

$\longrightarrow \sf \: S _{n} = \Sigma _n = 1 \: to \: 11th \: t _n \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf S _{11} = \Sigma _n = 1 \: to \: 11th \bigg[n(2n - 1) \bigg] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf S _{n} = \Sigma _n = 1 \: to \: 11th \bigg[2 {n}^{2} - n\bigg] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf S _{n} = \dfrac{\bigg[2n(n + 1)(2n + 1)\bigg]}{6} \: - \dfrac{\bigg[n(n + 1)\bigg]}{2} \\ \\ \\ \longrightarrow \sf S _{11} = \dfrac{\bigg[2 \times 11(12)(23)\bigg]}{6} \: - \dfrac{\bigg[11(12)\bigg]}{2}\qquad \\ \\ \\ \longrightarrow \sf S _{11} =1012 - 66\qquad\qquad\qquad\qquad\qquad\: \: \: \: \: \: \\ \\ \\ \longrightarrow {\pink{\underline{\boxed{\pmb{\tt{ S _{11} =946}}}}\purple\bigstar}}\qquad\qquad\qquad\qquad \qquad\: \: \: \: \: \: \: \: \:$

Therefore:

• The sum of the 11th terms is 946

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