Question

Write the value of nth bracket in
(1)+(1+3)+(1+3+5)+........
a) n? b) n(n+1) C) n(n+1)/2 d) (3n-2) (4n-1)​

Answer 1

Answer:

$n {2}$

is the answer........

Answer 2

Answer:

Value of nth bracket in (1)+(1+3)+(1+3+5)+........ = n²

Step-by-step explanation:

Given,

(1)+(1+3)+(1+3+5)+........

Required to find,

The value of the nth term (1)+(1+3)+(1+3+5)+........

Recall the formula

The sum to n-terms of an AP, $S_n$= $\frac{n}{2}[2a+(n-1)d]$ --------------(1)

where 'a' is the first term and 'd' is the common difference of AP

Solution:

The nth term of this series 1+3+5+.......(nterms)

Here, 1,3,5,............. form an AP with first term = 1 and common difference = 2

1+3+5+.......  =  Sum to 'n' terms of the AP, where a = 1 and d= 2

Substituting the value of 'a' and 'd' in equation (1)

1+3+5+...... =  $\frac{n}{2}[2 X 1+(n-1)2]$

= $\frac{n}{2}[2+2n-2]$

=$\frac{n}{2}[2n]$

=n²

1+3+5+.......  ( nterms)  = n²

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