Sum of first n natural numbers

Total saving is (S_{49}=\frac{49\times50}{2}=1225) rupees. A regular increase of (1) gives (S_n).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(S_{21}=\frac{21\times22}{2}=231\). Therefore, the required value is \(4\times231=924\). A choice such as \(904\) may result from an error while finding the sum or multiplying it. Exam tip: calculate \(n(n+1)/2\) first, then multiply by the given factor.

Among two consecutive numbers, one is even and the other is odd. Thus \(n(n+1)\) always has a factor of 2, so division by 2 gives an integer sum. Exam tip: use the even-odd pattern for consecutive integers.

The sum of the first 39 natural numbers is \(S_{39}=\frac{39\times40}{2}=780\). Therefore, the required difference is \(780-39=741\). Hence, 741 is the correct option. Subtracting 39 removes the last term, so the result is also the sum of the first 38 natural numbers, \(S_{38}\). Exam tip: First use \(\frac{n(n+1)}{2}\) to find the sum, then perform the required subtraction.

There are \(34-23+1=12\) terms from 23 to 34. This is an arithmetic progression, so its sum is \(\frac{12}{2}(23+34)=6\times57=342\). Therefore, 342 is correct. A value such as 352 can result from an error while counting the terms or finding the average. Exam tip: for consecutive numbers, add the first and last terms and multiply by half the number of terms.

For the sum of the first n natural numbers, \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_n-S_{n-1}=n\), since the difference leaves only the newly added term n. Given \(S_n-S_{n-1}=62\), we get \(n=62\). If n were 61, the difference would be 61, not 62. Exam tip: The difference between two consecutive partial sums is always the newly added term of the sequence.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=75\), we get \(\frac{75\times 76}{2}=75\times 38=2850\). The value \(2775\) is the sum of the first 74 natural numbers, so it is a close but incorrect option. Exam tip: In the formula, remember to use \(n+1\) along with \(n\).

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. To obtain \(S_{22}\) from \(S_{18}\), add the next four natural numbers, \(19,20,21\), and \(22\): \(S_{22}=171+19+20+21+22=253\). Hence, the correct answer is 253. One must include every term from 19 through 22, not skip any term. Exam tip: When moving from \(S_m\) to \(S_n\), add all terms from \(m+1\) to \(n\).

If \(S_{45}=1+2+\cdots+45\), then adding \(46\), \(47\), and \(48\) gives \(1+2+\cdots+45+46+47+48=S_{48}\). \(S_{47}\) would include terms only up to 47; adding 48 makes the result \(S_{48}\). Exam tip: when consecutive next natural numbers are added, the subscript of the sum becomes the last number added.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{30}=\frac{30\times31}{2}=465\) and \(S_{15}=\frac{15\times16}{2}=120\). Therefore, \(S_{30}:S_{15}=465:120=31:8\). The ratio \(31:9\) would result from using an incorrect value for the second sum. Exam tip: simplify a ratio at the end by dividing both terms by their greatest common factor.

Total flowers are \(S_{28}=\frac{28\times29}{2}=406\). Triangular rows indicate \(S_n\).

The difference of two consecutive sums is the new term, so (S_{64}-S_{63}=64). You can answer without finding full sums.

The sum of the first \(t\) natural numbers is \(\frac{t(t+1)}{2}\). Thus, \(\frac{t(t+1)}{2}=231\), so \(t(t+1)=462\). Since \(21\times22=462\), \(t=21\). Therefore, \(t+5=21+5=26\). Option 25 would result from taking \(t=20\), but the sum from 1 to 20 is 210. Exam tip: When a sum is given, first find \(n\) using \(\frac{n(n+1)}{2}\), then evaluate the expression asked.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{68}=\frac{68\times69}{2}=34\times69=2346\). Hence, 2346 is correct. A value such as 2356 can result from an error in multiplication or addition. Exam tip: when \(n\) is even, divide \(n\) by 2 first to calculate quickly.

(S_{13}=91), (S_{16}=136), and (S_{20}=210), so the total is (437). Write the three sums separately and add.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{a(a+1)}{2}=820\), we get \(a=40\), since \(\frac{40\times41}{2}=820\). Similarly, \(\frac{b(b+1)}{2}=1035\) gives \(b=45\), since \(\frac{45\times46}{2}=1035\). Therefore, \(b-a=45-40=5\). Option 4 may seem close, but it does not equal the actual difference between the indices. Exam tip: equate the given sum to \(\frac{n(n+1)}{2}\) to find the index n.

This becomes (S_{55}), and (S_{55}=1540). Adding consecutive next terms makes the new last term (n).

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{80}=\frac{80\times81}{2}=40\times81=3240\). Hence, option B is correct. The other values result from an error in multiplying 80 and 81 or dividing by 2. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of consecutive natural numbers starting from 1.

After the first n numbers, the next natural number is \(n+1\). Hence \(S_{n+1}=S_n+(n+1)\), so the difference is \(n+1\), not n. Exam tip: compare consecutive sums by identifying the newly added term.

(S_{25}=325) and (S_{27}=378), so the sum is (703). Do not make a mistake with nearby indices.