What-s-the-sum-of-the-odd-numbers-between-2313-and-4718-

Question Number 179773 by Acem last updated on 02/Nov/22

{W h a t}^{'} s t h e s u m o f t h e o d d n u m b e r s

b e t w e e n 2313 a n d 4718

Commented by CElcedricjunior last updated on 05/Nov/22

c a l c u l o n s l a s o m m e d e s n o m b r e s

c o m p r i s e s e n t r e 2313 e t 4718

\Leftrightarrow 2314 + 2315 + \dots \dots + 4717

= (2313 + 1) + (2313 + 2) + \dots \dots (2313 + 2404)

= 2313 \times 2404 + \frac{2404 \times 2405}{2}

= 2313 \times 2404 + 1202 \times 2405

= 8.451.262

d^{'} o \overset{`}{u} \underset{k = 2314}{\sum^{4717}} k = 8.451.262

Answered by Rasheed.Sindhi last updated on 02/Nov/22

S u m o f o d d n u m b e r s

b e t w e e n 2313 a n d 4718 (e x c l u s i v e l y)

2315 + 2317 + \dots + 4717

a = 2315, l = 4717, d = 2,

l = a_{n} = a + (n - 1) d = 2315 + 2 (n - 1) = 4717

n = (4717 - 2315 + 2) / 2 = 1202

\begin{matrix} S_{n} = \frac{n}{2} (a + l) \end{matrix}

S_{1202} = \frac{1202}{2} (2315 + 4717) = 4226232

Commented by Acem last updated on 02/Nov/22

T h a n k s f o r y o u r e f f o r t s a n d w r i t i n g d e t a i l s

F o r m e i u s e t h e s u m t i l l 1 a n d s u b t r a c t

S u m = {(\frac{n_{1}}{2})}^{2} - {(\frac{n_{2}}{2})}^{2}; n_{i} e v e n

= {(\frac{4718}{2})}^{2} - {(\frac{2314}{2})}^{2} = 4 226 232

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