Question

The sum of n terms of 2, 4, 6, 8.... is
(a) n(n+1)
(b) (1/2)(n+1)
(c) nin-1)​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$Option \green {(a)} \: is \: correct$

Step-by-step explanation:

$Given \: sequence : 2,4,6,8,\cdot\cdot\cdot\\n\:terms \:is \:an \:A.P$

$First \:term (a) = 2$

$Common \: difference (d) = a_{2} - a_{1}\\=4 -2 \\= 2$

$\boxed {\pink { Sum \: of \: n\:terms (S_{n}) = \frac{n}{2}[2a+(n-1)d]}}$

$S_{n} = \frac{n}{2}[ 2\times 2 + (n-1)\times 2]$

$= \frac{n}{2}[4+2n-2]$

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

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

$\green {=n(n+1)}$

Therefore.,

$\red {The\:sum \:of\:n\:terms \:of \:2,4,6,8,\cdot\cdot\cdot }\green {=n(n+1)}$

$Option \green {(a)} \: is \: correct$

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