a-n-number-series-a-1-5-d-3-a-2-2-a-1-2-a-4-2-a-3-2-a-6-2-a-5-2-a-10-2-a-9-2-

Question Number 207262 by hardmath last updated on 10/May/24

a_{n} - number series

a_{1} = 5

d = 3

a_{2}^{2} - a_{1}^{2} + a_{4}^{2} - a_{3}^{2} + a_{6}^{2} - a_{5}^{2} + \dots + a_{10}^{2} - a_{9}^{2} = ?

Answered by A5T last updated on 10/May/24

(a_{2} - a_{1}) (a_{2} + a_{1}) = d \times [2 a_{1} + d]

a_{2 n}^{2} - a_{2 n - 1}^{2} = (a_{2 n} - a_{2 n - 1}) (a_{2 n} + a_{2 n - 1})

= d \times [a_{1} + (2 n - 1) d + a_{1} + (2 n - 2) d]

= d \times [2 a_{1} + d (4 n - 3)]

\underset{n = 1}{\sum^{5}} a_{2 n}^{2} - a_{2 n - 1}^{2} = d [10 a_{1} + d (1 + 5 + 9 + 13 + 17)]

= 3 [50 + 3 (45)] = 555

Commented by hardmath last updated on 10/May/24

perfect dear professor thankyou