1-2-4-7-11-16-up-to-25-terms-

Question Number 202780 by BaliramKumar last updated on 03/Jan/24

1 + 2 + 4 + 7 + 11 + 16 + \dots \dots \dots \dots up to 25 terms = ?

Answered by mr W last updated on 03/Jan/24

a_{1} = 1

a_{2} = a_{1} + 1

a_{3} = a_{2} + 2

a_{4} = a_{3} + 3

\dots

a_{n} = a_{n - 1} + (n - 1)

Σ :

a_{n} = 1 + 1 + 2 + 3 + \dots + (n - 1) = 1 + \frac{n (n - 1)}{2}

S_{n} = \underset{k = 1}{\sum^{n}} a_{k} = n + \frac{n (n + 1) (2 n + 1)}{12} - \frac{n (n + 1)}{4}

\Rightarrow S_{n} = \frac{n (n^{2} + 5)}{6}

w i t h n = 25 :

a_{25} = 1 + \frac{25 \times 24}{2} = 301

S_{25} = \frac{25 (25^{2} + 5)}{6} = 2625

i . e .

1 + 2 + 4 + 7 + 11 + 16 + \dots + 301 = 2625

Commented by BaliramKumar last updated on 03/Jan/24

Thanks Sir

Answered by talminator2856792 last updated on 03/Jan/24

1 + 2 + 3 + 4 + \dots + n

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

1 + 3 + 6 + 10 + \dots + \frac{n (n + 1)}{2}

= \frac{n (n + 1) (n + 2)}{6}

1 + 2 + 4 + 7 + 11 + \dots + a_{n}

= (1 + 2 + 3 + 4 + 5 + \dots + n) +

(0 + 0 + 1 + 3 + 6 + \dots + \frac{(n - 1) (n - 2)}{2})

= \frac{n (n + 1)}{2} + \frac{n (n - 1) (n - 2)}{6}

\to 1 + 2 + 4 + 7 + 11 + 16 + \dots + a_{25}

= 325 + 2300

= 2625

Commented by BaliramKumar last updated on 03/Jan/24

Thanks Sir