1.

Find the sum of the series `1+ (1+2) + (1+2+3) + (1+2 + 3 +4)+ … + (1+2+3+… + 20)`.A. 1470B. 1540C. 1610D. 1370

Answer» Correct Answer - B
`1+(1+2)+ (1+2+3)+(1+2+3+4) +...20` terms.
`= 1+3+6+10 + 15 + 21 + 28 + 36 +m 45+55 + 66 + 78 + 91 + 105+ 120 + 136 + 153 + 171 + 190 + 210`
`= 1540`
Alternate method :
`1+(1+2) + (1+2 + 3) + ...n`th term of the series is
`1+2 +3 + 4 + ...n = (n(n+1))/(2)`
Sum of `n` terms of the series
`= sum t_n`
`= sum (n(n+1))/(2) rArr = (1)/(2) sum (n^(2) + n)`
` = (1)/(2) ( sum n^(2) + sumn)`
`= (1)/(2)[(n(n+1) (2n+1))/(6) + (n(n+1))/(2)]`
`= (n(n+1))/(4) ((2n+1)/(3) +1)`
`rArr (n(n+1)(2n+4))/(12) = (n(n+1)(n+2))/(6)`
If `n= 20`
`sum t_(20) = (20(21)(22))/(6)`
`rArr (10xx 21xx 22)/(3) = 1540`.


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