Sum of first n natural numbers

(S_{64}=2080), (S_{84}=3570), and (S_{74}=2775), so the value is (2875). In mixed sums, write each (S_n) separately.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(S_{5p}=5050\), we get \(\frac{5p(5p+1)}{2}=5050\). Since \(\frac{100\times101}{2}=5050\), \(5p=100\), so \(p=20\). If p were 18, the index would be 90, whose sum is not 5050. Exam tip: recognise 5050 as half the product of two consecutive numbers first.

(S_{100}=5050) and (2S_{50}=2550), so the value is (2500). Keep the order of multiplication and subtraction clear.

The governing concept is the sum of the first n natural numbers. Since row r contains 4r lamps, the numbers of lamps in rows 1 through 35 are 4, 8, 12, and so on, up to 4×35. Factor out 4 from the total: total lamps = 4(1+2+3+…+35). Using the formula S_n = n(n+1)/2, we get S_35 = 35×36/2 = 630. Therefore, the total number of lamps is 4×630 = 2520. Hence option D is correct. The other choices do not result from applying the natural-number sum formula and multiplying by the four lamps-per-row factor.

(S_{104}=\frac{104\times105}{2}=5460), so (n=104). Match (2S_n) with the product of consecutive numbers.

The expression is (n+(n-4)=2n-4), and (2n-4=150) gives (n=77). Convert differences of consecutive sums into terms.

Since \(S_{75}=S_{68}+(69+70+\cdots+75)\), the required number is the sum of the terms from 69 to 75. There are 7 terms, so their sum is \(\frac{7}{2}(69+75)=\frac{7}{2}\times144=504\). Hence, 504 is correct. Exam tip: When finding \(S_b-S_a\), add the terms from \(a+1\) to \(b\).

(S_{53}=1431) and (S_{81}=3321), so (u+v=134). Identify the indices of both triangular numbers.

(S_{96}=4656), and (12.5%=\frac{1}{8}), so the value is (582). Converting percent to a fraction makes it easier.

(S_{70}=2485), (S_{71}=2556), and (S_{69}=2415), so the value is (2626). Subtract carefully with nearby indices.

The difference is ((n-5)+(n-4)+\cdots+n=6n-15), and (6n-15=621) gives (n=106). Form the sum of the last six terms.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=175\), we get \(\frac{175\times176}{2}=175\times88=15400\). Therefore, option B is correct. Option A can result from an error in multiplication or halving. Exam tip: always remember to use \(n+1\) in the formula.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers, so \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{140}=9870\), \(S_{130}=8515\), and \(S_{10}=55\). Therefore, \(S_{140}-S_{130}+S_{10}=9870-8515+55=1410\). Note that \(S_{140}-S_{130}\) is the sum of the numbers from 131 to 140; adding \(S_{10}\) gives 1410. Exam tip: You can also verify \(S_a-S_b\) by summing the terms from \(b+1\) to \(a\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(\frac{n(n+1)}{2}=1770\) gives \(n=59\), since \(\frac{59\times60}{2}=1770\). Therefore, \(2k+1=59\), so \(2k=58\) and \(k=29\). Option 30 may result from incorrectly taking the index as 60 instead of 59. Exam tip: first find n from \(S_n\), then equate the given index \(2k+1\) to n.

The sum is (S_{160}-S_{120}=12880-7260=5620). When starting from (121), subtract the sum up to (120).

(S_{80}=3240), so the difference is (81+82+\cdots+88=676). Find the sum of the next (8) terms separately.

(S_{199}=19900) and (S_{200}=20100), so it first exceeds (20000) at (n=200). Checking nearby values is the safest method.

Total vehicles are \(5S_{42}=5\times903=4515\). In a pattern like (5r), take (5) outside the sum.

Using \(S_n=\frac{n(n+1)}{2}\), we get \(8S_n+1=4n(n+1)+1=(2n+1)^2\), a perfect square. However, \(S_n\) itself is not always odd or a square. Exam tip: substitute the standard sum formula to test such identities.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers, so \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{36}=666\), \(S_{72}=2628\), and \(S_{18}=171\). Therefore, \(S_{36}+S_{72}-S_{18}=666+2628-171=3123\). The option 3143 can result from an error of 20 while subtracting. Exam tip: calculate each \(S_n\) separately before performing the final addition and subtraction.