Special Series

JEE Main Mathematics · Progressions · 16 questions, latest first

• The sum 1+1/2(1^2+2^2)+1/3(1^2+2^2+3^2)+…. upto 10 terms is equal to:20266 April, Shift 2 · Q6
• The value of 1^3-2^3+3^3-…+15^3 is:20266 April, Shift 1 · Q5
• The sum 1^3/1+(1^3+2^3)/(1+3)+(1^3+2^3+3^3)/(1+3+5)+⋯ up to 8 terms, is:20262 April, Shift 2 · Q5
• The positive integer n, for which the solutions of the equation x(x+2)+(x+2)(x+4)+⋯+(x+2 n-2)(x+2 n)=(8 n)/3 are two consecutive…202621 January, Shift 2 · Q3
• If 1/1^4+1/2^4+1/3^4+… ∞=(π^4)/90, 1/1^4+1/3^4+1/5^4+… ∞=α, 1/2^4+1/4^4+1/6^4+… ∞=β, then (α)/(β) is equal to20258 April, Shift 2 · Q6
• 1+3+5^2+7+9^2+… upto 40 terms is equal to20254 April, Shift 1 · Q5
• lim_n → ∞ ((1^2-1)(n-1)+(2^2-2)(n-2)+⋯ ⋯+((n-1)^2-(n-1)) · 1)/((1^3+2^3+⋯ ⋯+n^3)-(1^2+2^2+⋯+n^2)) is equal to:20246 April, Shift 2 · Q9
• The value of (1 × 2^2+2 × 3^2+….+100 ×(101)^2)/(1^2 × 2+2^2 × 3+….+100^2 × 101) is20244 April, Shift 2 · Q5
• Let S_n be the sum to n -terms of an arithmetic progression 3,7,11, … …. If 40<(6/(n(n+1)) Σ_k=1^n S_k)<42, then n equals ____.202430 January, Shift 2 · Q25
• Let α=1^2+4^2+8^2+13^2+19^2+26^2+… upto 10 terms and β=Σ_n=1^10 n^4. If 4 α-β=55 k+40, then k is equal to ____.202430 January, Shift 1 · Q24
• Let [α] denote the greatest integer ≤ α. Then [√1]+[√2]+[√3]+…+[√120] is equal to ____202313 April, Shift 2 · Q24
• The sum to 20 terms of the series 2 · 2^2-3^2+2 · 4^2-5^2+2 · 6^2-… … is equal to ____.202313 April, Shift 1 · Q24
• Let S_K=(1+2+…+K)/K and Σ_j=1^n S_j^2=n/A(B n^2+C n+D), where A, B, C, D ∈ N and A has least value. Then20238 April, Shift 1 · Q9
• If gcd(m, n)=1 and 1^2-2^2+3^2-4^2+…..+(2021)^2-(2022)^2+(2023)^2=1012 m^2 n then m^2-n^2 is equal to20236 April, Shift 2 · Q10
• The sum 1^2-2 · 3^2+3 · 5^2-4 · 7^2+5 · 9^2-…+15 · 29^2 is ____.202331 January, Shift 2 · Q84
• If (1^3+2^3+3^3+… up to n terms)/(1 · 3+2 · 5+3 · 7+… up to n terms)=9/5, then the value of n is202324 January, Shift 2 · Q83