Question

$given \: that1 + 2 + 3 + ........n = \\ \\ \frac{n(n + 1)}{2 } \\ \\ \\ determine \: (i )\: 50 + 51 + 52 + .......100 \\ \\ (ii)2 + 4 + 6 + ...... + 100 \\ \\ (iii)1 + 3 + 5 + ..... + 99$
​

Answer 1

Answer:

(i)3775

(ii)2550

(iii)2500

Step-by-step explanation:

(i) 51 + 52 + 53 + ...........+ 100

 = (1 + 2 + 3 + ....... + 100) - (1 + 2 + 3 + ........ + 50)--------------------(1)

                                                                {∴Given: it is in the form of $\frac{n(n+1)}{2}$.

                                                                   So,  n₁=100 , n₂=50}

Now,

S₁=n₁/2(a+l)

S₂=n₂/2(a+l)                                                     (∴a=first term,l=last term)

In (1), we have to find S₁-S₂

S₁-S₂=[100/2(1+100)]-[50/2(50+1)]

S₁-S₂= (5050 - 1275) = 3775

(ii)Here, all the even numbers are taken from 1 to 100.

2+4+6+8+...+98+100  

On taking 2 common, we get

2(1+2+3+4+...+49+50)

We know the formula for sum of continuous series that is $\frac{n(n+1)}{2}$=2[n(n+1)2]

=n(n+1)  

=50(51)

⇒50*51=2550

(iii)The Common Difference is 3–1 = 2

Total no. of terms can be found using this formula:

aₙ=a+(n-1)d

aₙ is 99 (here).

d is the common difference.

Here, d=5-3=3-1=2

99 = 1+(n-1)2

⇒98= 2n-2

⇒2n = 100

⇒n = 50

There are total 50 terms in this series.

Sum= $\frac{n}{2} (a+l)$

S = 50/2(1+99)

S = 25*100

S = 2500

Answer 2

Answer:

Hey Buddy here's ur answer

(i) 51 + 52 + 53 + ...........+ 100

= (1 + 2 + 3 + ....... + 100) - (1 + 2 + 3 + ........ + 50)--------------------(1)

{∴Given: it is in the form of \frac{n(n+1)}{2}

2

n(n+1)

.

So, n₁=100 , n₂=50}

Now,

S₁=n₁/2(a+l)

S₂=n₂/2(a+l) (∴a=first term,l=last term)

In (1), we have to find S₁-S₂

S₁-S₂=[100/2(1+100)]-[50/2(50+1)]

S₁-S₂= (5050 - 1275) = 3775

(ii)Here, all the even numbers are taken from 1 to 100.

2+4+6+8+...+98+100

On taking 2 common, we get

2(1+2+3+4+...+49+50)

We know the formula for sum of continuous series that is \frac{n(n+1)}{2}

2

n(n+1)

=2[n(n+1)2]

=n(n+1)

=50(51)

⇒50*51=2550

(iii)The Common Difference is 3–1 = 2

Total no. of terms can be found using this formula:

aₙ=a+(n-1)d

aₙ is 99 (here).

d is the common difference.

Here, d=5-3=3-1=2

99 = 1+(n-1)2

⇒98= 2n-2

⇒2n = 100

⇒n = 50

There are total 50 terms in this series.

Sum= \frac{n}{2} (a+l)

2

n

(a+l)

S = 50/2(1+99)

S = 25*100

S = 2500