Question

if (2n+1)+(2n+3)+(2n+5)+....+(2n+47)=5280. find the value of 1+2+3+....+n

Answer 1

ANSWER:

• The value of 1 + 2 + 3 +....+ n = 4851

GIVEN:

• (2n + 1 ) + (2n + 3) + (2n + 5) +....+ (2n + 47) = 5280.

TO FIND:

• The value of 1 + 2 + 3 +....+ n.

EXPLANATION:

(2n + 1 ) + (2n + 3) + (2n + 5) +....+ (2n + 47) = 5280.

$\boxed { \bold{ \large{S_m= \dfrac{m}{2}(a + l)}}}$

Here m is the number of terms.

a = (2n + 1 )

l = (2n + 47)

$\sf S_m = 5280$

$\boxed {\large{ \bold{ m= \dfrac{l-a}{d} + 1}}}$

d = t₂ - t₁ = (2n + 3) - (2n + 1 )

d = 3 - 1 = 2

$\sf{m= \dfrac{2n + 47 -(2n + 1 )}{2} + 1}$

$\sf{m= \dfrac{2n + 47 -2n - 1 }{2} + 1}$

$\sf{m= \dfrac{46}{2} + 1}$

$\sf{m = 23 + 1}$

$\sf m = 24$

$\sf{5280= \dfrac{24}{2}(2n +1 + 2n + 47)}$

$\sf{5280= 12(4n + 48)}$

$\sf{440= 4n + 48}$

$\sf{110= n + 12}$

$\sf{n = 98}$

Substitute n = 98 in 1 + 2 + 3 +....+ n.

1 + 2 + 3 +....+ 98

$\boxed { \bold{ \large{Sum= \dfrac{m(m + 1)}{2}}}}$

Here m is the number of terms.

m = 98

$\sf{Sum= \dfrac{98(98 + 1)}{2}}$

$\sf{Sum= 49(99)}$

$\sf{Sum= 4851}$

Hence the value of 1 + 2 + 3 +.... + n = 4851