Exams Joint Entrance Examination Mathematics Sequences and Series Question 110 Question 110 of 519 Medium NUMERICAL
The sum to 20 20 20 terms of the series \[2 \cdot 2^{2}-3^{2}+2 \cdot 4^{2}-5^{2}+2 \cdot 6^{2}-\ldots \ldots\] is equal to \_\_\_\_\_\_\_\_\_\_.
Answer: 1310
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∑ r = 1 10 ( 2 ⋅ ( 2 r ) 2 − ( 2 r + 1 ) 2 ) = ∑ r = 1 10 ( 8 r 2 − 4 r 2 − 4 r − 1 ) = ∑ r = 1 10 ( 4 r 2 − 4 r − 1 ) = 4 ⋅ 10 ⋅ 11 ⋅ 21 6 − 4 10 ⋅ 11 2 − 10 = 44 ⋅ 35 − 220 − 10 = 1540 − 230 = 1310 \begin{gathered} \sum _{r=1}^{10}(2\cdot (2r{)}^{2}-(2r+1{)}^{2}) \\ =\sum _{r=1}^{10}(8{r}^{2}-4{r}^{2}-4r-1) \\ =\sum _{r=1}^{10}(4{r}^{2}-4r-1) \\ =\frac{4\cdot 10\cdot 11\cdot 21}{6}-4\frac{10\cdot 11}{2}-10 \\ =44\cdot 35-220-10 \\ =1540-230=1310 \end{gathered} r = 1 ∑ 10 ( 2 ⋅ ( 2 r ) 2 − ( 2 r + 1 ) 2 ) = r = 1 ∑ 10 ( 8 r 2 − 4 r 2 − 4 r − 1 ) = r = 1 ∑ 10 ( 4 r 2 − 4 r − 1 ) = 6 4 ⋅ 10 ⋅ 11 ⋅ 21 − 4 2 10 ⋅ 11 − 10 = 44 ⋅ 35 − 220 − 10 = 1540 − 230 = 1310 Answer the question to unlock