Question

Calculati urmatoarele sume 1+2+3...+80. 2+4+6....+100. 1+3+5+...+99. 3+7+11+15+...+43

Answer 1

(a)

1+2+3+..........+80

we can see that this is arithematic sequence

so, first term is

$a_1=1$

common difference is

$d=2-1$

$d=1$

total term is

$n=80$

now, we can find sum of nth term

$S_n=\frac{n}{2}(2a_1+(n-1)d)$

now, we can plug values

and we get

$S_8_0=\frac{80}{2}(2*1+(80-1)*1)$

$S_8_0=3240$...........Answer

(B)

2+4+6+.......+100

we can see that this is arithematic sequence

so, first term is

$a_1=2$

common difference is

$d=4-2$

$d=2$

total term is

$a_n=a_1+(n-1)d$

$100=2+(n-1)*2$

$98=(n-1)*2$

$n=50$

now, we can find sum of nth term

$S_n=\frac{n}{2}(2a_1+(n-1)d)$

now, we can plug values

and we get

$S_5_0=\frac{50}{2}(2*2+(50-1)*2)$

$S_5_0=2550$..........Answer

(C)

1+3+5+...+99

we can see that this is arithematic sequence

so, first term is

$a_1=1$

common difference is

$d=3-1$

$d=2$

total term is

$a_n=a_1+(n-1)d$

$99=1+(n-1)*2$

$98=(n-1)*2$

$n=50$

now, we can find sum of nth term

$S_n=\frac{n}{2}(2a_1+(n-1)d)$

now, we can plug values

and we get

$S_5_0=\frac{50}{2}(2*1+(50-1)*2)$

$S_5_0=2500$..........Answer

(D)

3+7+11+15+...+43

we can see that this is arithematic sequence

so, first term is

$a_1=3$

common difference is

$d=7-3$

$d=4$

total term is

$a_n=a_1+(n-1)d$

$43=3+(n-1)*4$

$40=(n-1)*4$

$n=11$

now, we can find sum of nth term

$S_n=\frac{n}{2}(2a_1+(n-1)d)$

now, we can plug values

and we get

$S_1_1=\frac{11}{2}(2*3+(11-1)*4)$

$S_1_1=253$..........Answer