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Grade: 12th pass

                        

Sum of the first 15 terms of the series 2 2 +5 2 +8 2 +....... is options are 10415 10425 10440 10455

4 years ago

Answers : (1)

Nishant Vora
IIT Patna
askIITians Faculty
2467 Points
							Hi,

Please find the solution

22+52+82+......15 terms
=22+52+82+.....+442
=\sum_{n=0}^{14} (2 + 3n)^2
=\sum_{n=0}^{14} (4 + 9n^2 + 12n)
=\sum_{n=0}^{14} 4 + 9\sum_{n=0}^{14}n^2 + 12\sum_{n=0}^{14}n
=4\times 15 + 9 \frac{14(14+1)(2*14+1)}{6} + 12 \frac{14*(14+1)}{2}

Now solve this



4 years ago
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