Question

1+(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5+)+....+(1+2+3+4+....90)=? Find the sum of the series

Answer 1

Answer:

1540

Step-by-step explanation:

1+(1+2)+ (1+2+3)+(1+2+3+4) +...20

terms. <br>

= 1+3+6+10 + 15 + 21 + 28 + 36 +m 45+55 + 66 + 78 + 91 + 105+ 120 + 136 + 153 + 171 + 190 + 210

<br>

= 1540

<br> Alternate method : <br>

1+(1+2) + (1+2 + 3) + ...n

th term of the series is <br>

1+2 +3 + 4 + ...n = (n(n+1))/(2)

<br> Sum of

n

terms of the series <br>

= sum t_n

<br>

= sum (n(n+1))/(2) rArr = (1)/(2) sum (n^(2) + n)

<br>

= (1)/(2) ( sum n^(2) + sumn)

<br>

= (1)/(2)[(n(n+1) (2n+1))/(6) + (n(n+1))/(2)]

<br>

= (n(n+1))/(4) ((2n+1)/(3) +1)

<br>

rArr (n(n+1)(2n+4))/(12) = (n(n+1)(n+2))/(6)

<br> If

n= 20

<br>

sum t_(20) = (20(21)(22))/(6)

<br>

rArr (10xx 21xx 22)/(3) = 1540