Sum of first n natural numbers

(S_n-S_{n-3}=(n-2)+(n-1)+n=3n-3), so (3n-3=144) gives (n=49). Write the difference as the last three terms.

Total blocks are (S_r=1275), and (S_{50}=1275), so there are (50) rows. In such figures, the total objects form a triangular number.

The sum of the first n natural numbers is
\(S_n=\frac{n(n+1)}{2}\). From
\(\frac{a(a+1)}{2}=1540\), we get
\(a=55\), and from
\(\frac{b(b+1)}{2}=2080\), we get
\(b=64\). Therefore,
\(b-a=64-55=9\). Although 8 may seem close, the indices corresponding to the two given sums differ by exactly 9. Exam tip: equate the given sum to
\(\frac{n(n+1)}{2}\) and solve for n.

This is an arithmetic sequence with first term 46, last term 90, and \(90-46+1=45\) terms. Therefore, \(S=\frac{45}{2}(46+90)=\frac{45}{2}\times136=3060\). Option 4095 is the sum from \(1\) to \(90\), so it incorrectly includes the terms from \(1\) to \(45\). Exam tip: while counting consecutive integers from one endpoint to another, always add \(1\).

(S_{m+1}-S_m=m+1), so (m+1=76) and (m=75). The difference of consecutive sums is the next term.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{100}=\frac{100\times101}{2}=5050\) and \(S_{75}=\frac{75\times76}{2}=2850\). The required difference is \(5050-2850=2200\), so option B is correct. It is also the sum of the integers from 76 to 100. Option 2250 can result from an incorrect value of \(S_{75}\) or a subtraction error. Exam tip: interpret \(S_b-S_a\) as the sum of terms from \(a+1\) to \(b\).

(S_{76}=2926) and (S_{77}=3003), so the sum first exceeds (3000) at (n=77). In boundary questions, check the two nearby sums.

(S_{25}=325), (S_{35}=630), and (S_{45}=1035), so the total is (1990). Write all three sums separately and add.

For the sum of the first n natural numbers, \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=3240\), we get \(n=80\). Therefore, \(S_{n-4}=S_{76}\). Now \(S_{76}=S_{80}-(77+78+79+80)=3240-314=2926\). Option 2916 results from subtracting the last four terms incorrectly. Exam tip: To find \(S_{n-k}\), subtract the last k terms from \(S_n\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). For \(n=96\), \(\frac{96\times97}{2}=48\times97=4656\). Hence, option A is correct. A value such as \(4626\) may result from an error in multiplication or addition. Exam tip: for the sum from 1 to \(n\), use \(\frac{n(n+1)}{2}\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{40}=\frac{40\times41}{2}=820\), \(S_{60}=\frac{60\times61}{2}=1830\), and \(S_{50}=\frac{50\times51}{2}=1275\). Therefore, \(S_{40}+S_{60}-S_{50}=820+1830-1275=1375\). An answer such as 1385 can result from an error while evaluating one of the sums. Exam tip: calculate each \(S_n\) separately before performing the final addition and subtraction.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(S_{2p}=820\), so \(\frac{2p(2p+1)}{2}=820\). Since \(\frac{40\times41}{2}=820\), we get \(2p=40\), hence \(p=20\). If \(p=19\), the index would be \(38\), whose sum is not 820. Exam tip: When a sum is given, equate it to \(n(n+1)/2\) and find the index \(n\) first.

(S_{70}=2485) and (2S_{35}=1260), so the value is (1225). Keep the order of multiplication and subtraction clear.

Total lamps are (3S_{30}=3\times465=1395). If there is a common multiplier, it can be placed outside the sum.

(S_{90}=\frac{90\times91}{2}=4095), so (n=90). Matching (2S_n) with a consecutive product is useful.

The expression is (n+(n-2)=2n-2), so (2n-2=99) gives no natural (n). Check parity before choosing an option.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. Therefore, \(S_{50}-S_{44}\) contains only the terms from 45 to 50: \(45+46+47+48+49+50=285\). Hence, adding \(285\) to \(S_{44}\) gives \(S_{50}\). An option such as \(279\) is incorrect because it is not the complete sum of these six consecutive terms. Exam tip: To find \(S_b-S_a\), add the terms from \(a+1\) to \(b\).

(S_{49}=1225) and (S_{72}=2628), so (u+v=121). First identify the indices of both triangular numbers.

(S_{88}=3916), and (25%), or (\frac{1}{4}), of it is (979). Converting percent to a fraction makes it easier.

(S_{58}=1711), (S_{59}=1770), and (S_{57}=1653), so the value is (1887). Subtract carefully with nearby indices.