Question

find the sum 1.3+5.7+9.11+....up to n terms

Answer 1

Here is your answer friends , Hope it helps , if there is any mistake in formulas , forgive me and correct me

Answer 2

Answer:

Required sum is $\frac{n(n + 1)(2n + 7)}{6}$where n= odd integer.

Step-by-step explanation:

In the given series every term has two factors,If we write all the first factors of each terms in the same order.

Here give series is 1.3+5.7+9.11+....up to n terms.

$t_n = n(n + 2) = {n}^{2} + 2n$

Putting n=1,

$t_1 = 1(1+ 2) = {1}^{2} + 2.1$

Putting n=2,

$t_2 = {2}^{2} + 2 \times 2$

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$t_n = {n}^{2} + 2n$

We are adding each term upto $t_n$

Sum

$= ( {1}^{2} + {2}^{2} + .... {n}^{2} ) + 2(1 + 2 + ..... + n)$

$= \frac{n(n + 1)(2n + 1)}{6} + \frac{2n(n + 1)}{2}$

$= \frac{n(n + 1)}{2} ( \frac{2n + 1}{3} + 2) \\ =(\frac{n(n + 1)}{2})( \frac{2n + 7}{3} ) \\ = \frac{n(n + 1)(2n + 7)}{6}$

where n= odd integer.