Method of Differences

JEE Main Mathematics · Progressions · 22 questions, latest first

• If the sum of the first 10 terms of the series 1/(1+1^4 × 4)+2/(1+2^4 × 4)+3/(1+3^4 × 4)+4/(1+4^4 × 4)+… …. is m/n, gcd(m, n)=1,…20265 April, Shift 2 · Q5
• Σ_n=1^10(528/(n(n+1)(n+2))) is equal to:20265 April, Shift 1 · Q5
• If Σ_r=1^25(r/(r^4+r^2+1))=p/q, where p and q are positive integers such that gcd(p, q)=1, then p+q is equal to ____.202628 January, Shift 2 · Q22
• If the sum of the first 20 terms of the series (4 · 1)/(4+3 · 1^2+1^4)+(4 · 2)/(4+3 · 2^2+2^4)+(4 · 3)/(4+3 · 3^2+3^4)+(4 ·…20254 April, Shift 2 · Q5
• The sum 1+3+11+25+45+71+… upto 20 terms, is equal to20253 April, Shift 1 · Q7
• If the sum of the first 10 terms of the series (4 · 1)/(1+4 · 1^4)+(4 · 2)/(1+4 · 2^4)+(4 · 3)/(1+4 · 3^4)+….. is m/n, where…20252 April, Shift 2 · Q22
• The value of lim_n → ∞(Σ_k=1^n (k^3+6 k^2+11 k+5)/((k+3)!)) is:202529 January, Shift 1 · Q5
• For positive integers n, if 4 a_n=(n^2+5 n+6) and S_n=Σ_k=1^n(1/a_k), then the value of 507 S_2025 is:202528 January, Shift 2 · Q6
• Let S_n=1/2+1/6+1/12+1/20+… upto n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p…202524 January, Shift 1 · Q5
• If Σ_r=1^n T_r=((2 n-1)(2 n+1)(2 n+3)(2 n+5))/64, then lim_n → ∞ Σ_r=1^n(1/T_r) is equal to:202522 January, Shift 1 · Q5
• If (1/(α+1)+1/(α+2)+… …+1/(α+1012))-(1/(2 · 1)+1/(4 · 3)+1/(6 · 5)+… …+1/(2024 · 2023))=1/2024, then α is equal to ____.20249 April, Shift 2 · Q24
• If the sum of the series 1/(1 ·(1+d))+1/((1+d)(1+2 d))+…+1/((1+9 d)(1+10 d)) is equal to 5, then 50 d is equal to:20249 April, Shift 1 · Q5
• Let the first term of a series be T_1=6 and its r^th term T_r=3 T_r-1+6^r, r=2,3, …, n. If the sum of the first n terms of this…20246 April, Shift 1 · Q24
• If 1/(√1+√2)+1/(√2+√3)+…+1/(√99+√100)=m and 1/(1 · 2)+1/(2 · 3)+…+1/(99 · 100)=n, then the point (m, n) lies on the line20245 April, Shift 1 · Q6
• The sum of the series 1/(1-3 · 1^2+1^4)+2/(1-3 · 2^2+2^4)+3/(1-3 · 3^2+3^4)+… up to 10 -terms is202431 January, Shift 1 · Q5
• If S_n=4+11+21+34+50+… to n terms, then 1/60( S_29-S_9) is equal to202310 April, Shift 2 · Q8
• Let a_n be the n^th term of the series 5+8+14+23+35+50+… and S_n=Σ_k=1^n a_k. Then S_30-a_40 is equal to20238 April, Shift 2 · Q8
• The sum of the first 20 terms of the series 5+11+19+29+41+… is20236 April, Shift 1 · Q7
• Let a_1, a_2, a_3, …, a_n be n positive consecutive terms of an arithmetic progression. If d>0 is its common difference, then…20236 April, Shift 1 · Q8
• The sum to 10 terms of the series 1/(1+1^2+1^4)+2/(1+2^2+2^4)+3/(1+3^2+3^4)+…. is20231 February, Shift 1 · Q68
• Let a_1, a_2, …, a_n be in A.P. If a_5=2 a_7 and a_11=18, then 12(1/(√(a_10)+√(a_11))+1/(√(a_11)+√(a_12))+…+1/(√(a_17)+√(a_18)))…202331 January, Shift 1 · Q85
• If a_n=-2/(4 n^2-16 n+15), then a_1+a_2+… …+a_25 is equal to:202330 January, Shift 1 · Q66