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If
1+2+3+
b)
(2,3,5,1,x
a) 46656

+n=36 then find 13 c)
(4//9,25,49,121)
b) 1296

$3$ . If $1+2+3+$ $\newline$ b) $(2,3,5,1, x$ $\newline$ a) $46656$ $\newline$ $+n=36$ then find $13$ c) $(4 / 9,25,49,121)$ $\newline$ b) $1296$

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Sum of finite series starts from 1

Full solution

Q.

3

. If

1+2+3+

\newline

b)

(2,3,5,1, x

\newline

a)

46656

\newline

+n=36

then find

13

c)

(4 / 9,25,49,121)

\newline

b)

1296

Set up equation: The sum of the first $n$ natural numbers is given by the formula $S = \frac{n(n+1)}{2}$ . We know $S = 36$ , so we can set up the equation $\frac{n(n+1)}{2} = 36$ .
Eliminate fraction: Multiply both sides by $2$ to get rid of the fraction: $n(n+1) = 72$ .
Solve quadratic equation: Now we need to solve the quadratic equation $n^2 + n - 72 = 0$ .
Factor equation: Factor the quadratic equation: n+\(9)(n $-8$ ) = $0$ \
Find valid solution: Set each factor equal to zero and solve for $n$ : $n+9 = 0$ or $n-8 = 0$ .
Find valid solution: Set each factor equal to zero and solve for $n$ : $n+9 = 0$ or $n-8 = 0$ .Solving $n+9 = 0$ gives $n = -9$ , which doesn't make sense for the sum of natural numbers, so we discard this solution.
Find valid solution: Set each factor equal to zero and solve for $n$ : $n+9 = 0$ or $n-8 = 0$ .Solving $n+9 = 0$ gives $n = -9$ , which doesn't make sense for the sum of natural numbers, so we discard this solution.Solving $n-8 = 0$ gives $n = 8$ , which is a valid solution.

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