Question below What is the partial sum?
\[\sum_{i=3}^{28}3+6i\]
split the sum, pull the constants, then work the variable formula
starting i at 1 is usually helpful in determining a count ... -2 the index and +2 the rule
\[\Large \sum_{i=3}^{28}3+6i\] \[\Large \sum_{i=3-2}^{28-2}3+6(i+2)\] \[\Large \sum_{i=1}^{26}3+6i+12\] \[\Large \sum_{i=1}^{26}15+6i\] \[\Large \sum_{i=1}^{26}15+\sum_{i=1}^{26}6i\] \[\Large 15(26)+6\sum_{i=1}^{26}i\]
Oh okay, would that leave me with 249 in the end?
maybe: 15(26) + 6(26)(27)/2
249__ missing a digit
216 249 282 312 These are the answers i have only triple digits
Oh, wow, it had been 8 not twenty eight sorry got them confused
lol, fine 15(6) + 6(6)(7)/2
\[\sum_{i=3}^{28}=3+6i = 3+6\times 3 + 3+6\times 4 + 3+6\times 5....3+6\times 28\] \[\sum_{i=3}^{28}=3+6i = 21 + 27 +33+....+171\] Here first trm a= 21 common difference d= 27-21=33-27=6 so this is an a sequence which is an A.P. ( Asthmatic Progression) tn= 171 since tn= a+(n-1)d 171=21+(n-1)6 171=21+6n-6 171=15+6n 171-15= 6n 156= 6n n= 156/6 n=26 Now using sum formula \[S_n = \frac{n}{2}(a+t_n) \] \[S_{26}= \frac{26}{2}(21+171) = 13(21+171)= 13 \times 192 =2496\]
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