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← JEE Main 2022 · 28 Jun · Shift 2
MathematicsInteger+4 / -1Question 25 of 90

JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 25

Let for n=1,2,…,50n=1,2,\dots,50, SnS_n be the sum of the infinite geometric progression whose first term is n2n^2 and whose common ratio is 1(n+1)2\dfrac1{(n+1)^2}. Then the value of 126+∑n=150(Sn+2n+1−n−1)\dfrac1{26}+\displaystyle\sum_{n=1}^{50}\left(S_n+\frac2{n+1}-n-1\right) is equal to ______.

Answer and worked solution

Official answer: 41651

Sum. Sn=n21−1(n+1)2=n(n+1)2n+2=n2+1−2n+2S_n=\dfrac{n^2}{1-\frac1{(n+1)^2}}=\dfrac{n(n+1)^2}{n+2}=n^2+1-\dfrac2{n+2}.

Term. Sn+2n+1−n−1=n2−n+2n+1−2n+2S_n+\frac2{n+1}-n-1=n^2-n+\frac2{n+1}-\frac2{n+2}.

Total. 42925−1275+2(12−152)=41651−12642925-1275+2\left(\frac12-\frac1{52}\right)=41651-\frac1{26}; adding 126\frac1{26} gives 41651.

Answer: 41651

Topics:Geometric progressionSpecial series sums
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