JEE Main 2023 · 13 April · Morning Shift · Mathematics · Sequences and Series
The sum to 202020 terms of the series 2⋅22−32+2⋅42−52+2⋅62−............2 \cdot {2}^{2} - {3}^{2} + 2 \cdot {4}^{2} - {5}^{2} + 2 \cdot {6}^{2} - . . . . . . . . . . . .2⋅22−32+2⋅42−52+2⋅62−............ is equal to__________ .
1310
Given,2⋅22−32+2⋅42−52+2⋅62−............2 \cdot {2}^{2} - {3}^{2} + 2 \cdot {4}^{2} - {5}^{2} + 2 \cdot {6}^{2} - . . . . . . . . . . . .2⋅22−32+2⋅42−52+2⋅62−............202020 terms,=∑r=110(2⋅(2r)2−(2r+1)2)= \mathop{∑}_{r = 1}^{10}\left(2 \cdot ( 2r{)}^{2} - ( 2r + 1{)}^{2}\right)=∑r=110(2⋅(2r)2−(2r+1)2)=∑r=110(8r2−4r2−4r−1)= \mathop{∑}_{r = 1}^{10}\left(8{r}^{2} - 4{r}^{2} - 4r - 1\right)=∑r=110(8r2−4r2−4r−1)=∑r=110(4r2−4r−1)= \mathop{∑}_{r = 1}^{10}\left(4{r}^{2} - 4r - 1\right)=∑r=110(4r2−4r−1)=4∑r=110r2−4∑r=110r−∑r=1101= 4\mathop{∑}_{r = 1}^{10}{r}^{2} - 4\mathop{∑}_{r = 1}^{10}r - \mathop{∑}_{r = 1}^{10}1=4∑r=110r2−4∑r=110r−∑r=1101=4⋅10⋅11⋅216−410⋅112−10= \dfrac{4 \cdot 10 \cdot 11 \cdot 21}{6} - 4\dfrac{10 \cdot 11}{2} - 10=64⋅10⋅11⋅21−4210⋅11−10=44⋅35−220−10= 44 \cdot 35 - 220 - 10=44⋅35−220−10=1540−230=1310= 1540 - 230 = 1310=1540−230=1310
Given,
2⋅22−32+2⋅42−52+2⋅62−............2 \cdot {2}^{2} - {3}^{2} + 2 \cdot {4}^{2} - {5}^{2} + 2 \cdot {6}^{2} - . . . . . . . . . . . .2⋅22−32+2⋅42−52+2⋅62−............202020 terms,
=∑r=110(2⋅(2r)2−(2r+1)2)= \mathop{∑}_{r = 1}^{10}\left(2 \cdot ( 2r{)}^{2} - ( 2r + 1{)}^{2}\right)=∑r=110(2⋅(2r)2−(2r+1)2)
=∑r=110(8r2−4r2−4r−1)= \mathop{∑}_{r = 1}^{10}\left(8{r}^{2} - 4{r}^{2} - 4r - 1\right)=∑r=110(8r2−4r2−4r−1)
=∑r=110(4r2−4r−1)= \mathop{∑}_{r = 1}^{10}\left(4{r}^{2} - 4r - 1\right)=∑r=110(4r2−4r−1)
=4∑r=110r2−4∑r=110r−∑r=1101= 4\mathop{∑}_{r = 1}^{10}{r}^{2} - 4\mathop{∑}_{r = 1}^{10}r - \mathop{∑}_{r = 1}^{10}1=4∑r=110r2−4∑r=110r−∑r=1101
=4⋅10⋅11⋅216−410⋅112−10= \dfrac{4 \cdot 10 \cdot 11 \cdot 21}{6} - 4\dfrac{10 \cdot 11}{2} - 10=64⋅10⋅11⋅21−4210⋅11−10
=44⋅35−220−10= 44 \cdot 35 - 220 - 10=44⋅35−220−10=1540−230=1310= 1540 - 230 = 1310=1540−230=1310
