Question

pratam 10 even number ka add nikalo​

Answer 1

Answer:

fwftstfafdafqfffwff se wfqf2eff

Answer 2

Answer:

hope it will help you

Step-by-step explanation:

If we take the sum of the first n positive integers, it will be n(n+1)2 .

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:S=1+2+⋯+(n−1)+n

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:S=1+2+⋯+(n−1)+n S=n+(n−1)+⋯+2+1

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:S=1+2+⋯+(n−1)+n S=n+(n−1)+⋯+2+1 Now, if we add up each column, we get

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:S=1+2+⋯+(n−1)+n S=n+(n−1)+⋯+2+1 Now, if we add up each column, we get2S=(n+1)+(n+1)+⋯+(n+1)+(n+1)

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:S=1+2+⋯+(n−1)+n S=n+(n−1)+⋯+2+1 Now, if we add up each column, we get2S=(n+1)+(n+1)+⋯+(n+1)+(n+1) 2S=n(n+1)

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:S=1+2+⋯+(n−1)+n S=n+(n−1)+⋯+2+1 Now, if we add up each column, we get2S=(n+1)+(n+1)+⋯+(n+1)+(n+1) 2S=n(n+1) S=n(n+1)2

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:S=1+2+⋯+(n−1)+n S=n+(n−1)+⋯+2+1 Now, if we add up each column, we get2S=(n+1)+(n+1)+⋯+(n+1)+(n+1) 2S=n(n+1) S=n(n+1)2 The only difference when calculating the sum of the first n even numbers is to multiply each term by 2 .

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:S=1+2+⋯+(n−1)+n S=n+(n−1)+⋯+2+1 Now, if we add up each column, we get2S=(n+1)+(n+1)+⋯+(n+1)+(n+1) 2S=n(n+1) S=n(n+1)2 The only difference when calculating the sum of the first n even numbers is to multiply each term by 2 .2(n(n+1)2)=n(n+1)

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:S=1+2+⋯+(n−1)+n S=n+(n−1)+⋯+2+1 Now, if we add up each column, we get2S=(n+1)+(n+1)+⋯+(n+1)+(n+1) 2S=n(n+1) S=n(n+1)2 The only difference when calculating the sum of the first n even numbers is to multiply each term by 2 .2(n(n+1)2)=n(n+1) Finally, substitute 10 for n to get

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:S=1+2+⋯+(n−1)+n S=n+(n−1)+⋯+2+1 Now, if we add up each column, we get2S=(n+1)+(n+1)+⋯+(n+1)+(n+1) 2S=n(n+1) S=n(n+1)2 The only difference when calculating the sum of the first n even numbers is to multiply each term by 2 .2(n(n+1)2)=n(n+1) Finally, substitute 10 for n to getn(n+1)=10(10+1)=110

If we take the sum of the first n positive integers, it will be n(n+1)2 .This is because if we write it out, we get this:S=1+2+⋯+(n−1)+n Writing it out another time, but this time backwards, will give you this:S=1+2+⋯+(n−1)+n S=n+(n−1)+⋯+2+1 Now, if we add up each column, we get2S=(n+1)+(n+1)+⋯+(n+1)+(n+1) 2S=n(n+1) S=n(n+1)2 The only difference when calculating the sum of the first n even numbers is to multiply each term by 2 .2(n(n+1)2)=n(n+1) Finally, substitute 10 for n to getn(n+1)=10(10+1)=110 The sum of the first 10 even numbers is 110 .

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