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The value of (2. 1P0 – 3. 2P1 + 4. 3P2 – .... up to 51th term) + (1! – 2! + 3! – ..... up to 51th term) is equal to : (1) 1 + (51)! (2) 1 – 51(51)! (3) 1 + (52)! (4) 1 |
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Answer» (3) 1 + (52)! S = (2. 1 p 0 – 3. 2 p 1 + 4. 3 p 2 ......... upto 51 terms) + (1! + 2! + 3! .......... upto 51 terms) [\(\therefore\) npn–1 = n!] \(\therefore\) S = (2 × 1! – 3 × 2! + 4 × 3! .... + 52.51!) + (1! – 2! + 3! ........... (51)!) = (2! – 3! + 4! ........ + 52!) + (1! – 2! + 3! – 4! + ...... + (51)!) = 1! + 52! |
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